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Let's consider the training dataset first. The data projected onto the principal axis (or axes) is principal component(s). In kPCA, eigenvectors of the kernel matrix directly give you principal components, but scaled to have unit sum-of-squares. To get the correct scaling, one needs to multiply them by the square roots of the respective eigenvalues, so $$\mathbf X_\mathrm{projected\:on\: axis \:\#i} = \sqrt{\lambda_i} \mathbf E_i,$$ where $\mathbf E_i$ and $\lambda_i$ are the $i$-th eigenvector and eigenvalue of the kernel matrix: $\mathbf K \mathbf E_i = \lambda_i \mathbf E_i$.

So the projection is not equal to $\mathbf K \mathbf E_i = \lambda_i \mathbf E_i$ as you wrote in your question, but to $\sqrt{\lambda_i} \mathbf E_i$.

This is easy to see by considering standard, non-kernel, PCA. There PCA amounts to an SVD decomposition of the (centered) data matrix: $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top,$$ where $\mathbf U \mathbf S$ are PCs (PC "scores") and $\mathbf V$ are principal axes. The kernel matrix is the so called Gram matrix $$\mathbf K = \mathbf X \mathbf X^\top = \mathbf U \mathbf S^2 \mathbf U^\top,$$ and so one can see that to get $\mathbf{US}$ its eigenvalues need to be square-rooted.

Turning now to your main question, you have a new (test) data point $\mathbf x$ (a row vector) that needs to be projected on the principal axes. When faced with a question about kPCA, always think about how you would do it in standard PCA. You need to compute $\mathbf x \mathbf V$. But say you don't know $\mathbf V$ (that's the case in kPCA). Well, you can compute $\mathbf k = \mathbf x \mathbf X^\top$, which is a (row) vector of kernels between the new data point and all the old ones. And now $$\mathbf x \mathbf V = \mathbf x \mathbf V \mathbf S \mathbf U^\top \mathbf U \mathbf S^{-1} = \mathbf x \mathbf X^\top \mathbf U \mathbf S^{-1} = \mathbf k \mathbf U \mathbf S^{-1}.$$ Rewriting this in your kPCA notation, $$\mathbf x_\mathrm{projected} = \mathbf k \mathbf E \boldsymbol \Lambda^{-1/2},$$ where $\boldsymbol \Lambda$ is the diagonal matrix with eigenvalues $\lambda_i$ on the diagonal.

Let's consider the training dataset first. Principal components (sometimes called PC "scores") are the centered data projected onto the principal axes. In kPCA, eigenvectors of the kernel matrix directly give you principal components, but scaled to have unit sum-of-squares. To get the correct scaling, one needs to multiply them by the square roots of the respective eigenvalues, so $$\mathbf X_\mathrm{projected\:on\: axis \:\#i} = \sqrt{\lambda_i} \mathbf E_i,$$ where $\mathbf E_i$ and $\lambda_i$ are the $i$-th eigenvector and eigenvalue of the kernel matrix: $\mathbf K \mathbf E_i = \lambda_i \mathbf E_i$. You wrote it wrong in your question.

This is easy to see by considering standard, non-kernel, PCA. Let $\mathbf X$ be the $n\times p$ centered data matrix. PCA amounts to an SVD decomposition of the (centered) data matrix: $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top,$$ where $\mathbf U \mathbf S$ are PCs (PC "scores") and $\mathbf V$ are principal axes. Usually PCA is introduced via eigen-decomposition of the covariance matrix: $\mathbf C = \mathbf X^\top\mathbf X/n$, which has $\mathbf V$ as eigenvectors. Alternatively, one can consider the so called Gram matrix $$\mathbf K = \mathbf X \mathbf X^\top = \mathbf U \mathbf S^2 \mathbf U^\top,$$ which has $\mathbf U$ as eigenvectors and $\mathbf S^2$ as eigenvalues. To get PCs $\mathbf{US}$ one needs to multiply eigenvectors $\mathbf U$ with the square roots of the eigenvalues.

The kernel matrix in kPCA is what I called Gram matrix above. So the bottomline is: multiply its eigenvectors with the square roots of its eigenvalues.

Turning now to your main question, you have a new (test) data point $\mathbf x$ (a row vector) that needs to be projected on the principal axes. When faced with a question about kPCA, always think about how you would do it in standard PCA. You need to compute $\mathbf x \mathbf V$. But say you don't know $\mathbf V$ (that's the case in kPCA). Well, you can compute $\mathbf k = \mathbf x \mathbf X^\top$, which is a (row) vector of kernels between the new data point and all the old ones. And now $$\mathbf x \mathbf V = \mathbf x \mathbf V \mathbf S \mathbf U^\top \mathbf U \mathbf S^{-1} = \mathbf x \mathbf X^\top \mathbf U \mathbf S^{-1} = \mathbf k \mathbf U \mathbf S^{-1}.$$ Rewriting this in your kPCA notation, $$\mathbf x_\mathrm{projected} = \mathbf k \mathbf E \boldsymbol \Lambda^{-1/2},$$ where $\boldsymbol \Lambda$ is the diagonal matrix with eigenvalues $\lambda_i$ on the diagonal.

Let's consider the training dataset first. The data projected onto the principal axis (or axes) is principal component(s). In kPCA, eigenvectors of the kernel matrix directly give you principal components, but scaled to have unit sum-of-squares. To get the correct scaling, one needs to multiply them by the square roots of the respective eigenvalues, so $$\mathbf X_\mathrm{projected\:on\: axis \:\#i} = \sqrt{\lambda_i} \mathbf E_i,$$ where $\mathbf E_i$ and $\lambda_i$ are the $i$-th eigenvector and eigenvalue of the kernel matrix (and not $\mathbf K \mathbf E_i = \lambda_i \mathbf E_i$ as you wrote in your question).

This is easy to see by considering standard, non-kernel, PCA. There PCA amounts to an SVD decomposition of the data matrix: $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top,$$ where $\mathbf U \mathbf S$ are PCs and $\mathbf V$ are principal axes. The kernel matrix is the so called Gram matrix $$\mathbf K = \mathbf X \mathbf X^\top = \mathbf U \mathbf S^2 \mathbf U^\top,$$ and so one can see that to get $\mathbf{US}$ its eigenvalues need to be square-rooted.

Turning now to your main question, you have a new (test) data point $\mathbf x$ (a row vector) that needs to be projected on the principal axes. When faced with a question about kPCA, always think about how you would do it in standard PCA. You need to compute $\mathbf x \mathbf V$. But say you don't know $\mathbf V$ (that's the case in kPCA). Well, you can compute $\mathbf k = \mathbf x \mathbf X^\top$, which is a (row) vector of kernels between the new data point and all the old ones. And now $$\mathbf x \mathbf V = \mathbf x \mathbf V \mathbf S \mathbf U^\top \mathbf U \mathbf S^{-1} = \mathbf x \mathbf X^\top \mathbf U \mathbf S^{-1} = \mathbf k \mathbf U \mathbf S^{-1}.$$ Rewriting this in your kPCA notation, $$\mathbf x_\mathrm{projected} = \mathbf k \mathbf E \boldsymbol \Lambda^{-1/2},$$ where $\boldsymbol \Lambda$ is the diagonal matrix with eigenvalues $\lambda_i$ on the diagonal.

Let's consider the training dataset first. The data projected onto the principal axis (or axes) is principal component(s). In kPCA, eigenvectors of the kernel matrix directly give you principal components, but scaled to have unit sum-of-squares. To get the correct scaling, one needs to multiply them by the square roots of the respective eigenvalues, so $$\mathbf X_\mathrm{projected\:on\: axis \:\#i} = \sqrt{\lambda_i} \mathbf E_i,$$ where $\mathbf E_i$ and $\lambda_i$ are the $i$-th eigenvector and eigenvalue of the kernel matrix: $\mathbf K \mathbf E_i = \lambda_i \mathbf E_i$.

So the projection is not equal to $\mathbf K \mathbf E_i = \lambda_i \mathbf E_i$ as you wrote in your question, but to $\sqrt{\lambda_i} \mathbf E_i$.

This is easy to see by considering standard, non-kernel, PCA. There PCA amounts to an SVD decomposition of the (centered) data matrix: $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top,$$ where $\mathbf U \mathbf S$ are PCs (PC "scores") and $\mathbf V$ are principal axes. The kernel matrix is the so called Gram matrix $$\mathbf K = \mathbf X \mathbf X^\top = \mathbf U \mathbf S^2 \mathbf U^\top,$$ and so one can see that to get $\mathbf{US}$ its eigenvalues need to be square-rooted.

Turning now to your main question, you have a new (test) data point $\mathbf x$ (a row vector) that needs to be projected on the principal axes. When faced with a question about kPCA, always think about how you would do it in standard PCA. You need to compute $\mathbf x \mathbf V$. But say you don't know $\mathbf V$ (that's the case in kPCA). Well, you can compute $\mathbf k = \mathbf x \mathbf X^\top$, which is a (row) vector of kernels between the new data point and all the old ones. And now $$\mathbf x \mathbf V = \mathbf x \mathbf V \mathbf S \mathbf U^\top \mathbf U \mathbf S^{-1} = \mathbf x \mathbf X^\top \mathbf U \mathbf S^{-1} = \mathbf k \mathbf U \mathbf S^{-1}.$$ Rewriting this in your kPCA notation, $$\mathbf x_\mathrm{projected} = \mathbf k \mathbf E \boldsymbol \Lambda^{-1/2},$$ where $\boldsymbol \Lambda$ is the diagonal matrix with eigenvalues $\lambda_i$ on the diagonal.

First, about the main dataset. The data projected onto the principal axis (or axes) is principal component(s). In kPCA, eigenvectors of the kernel matrix directly give you principal components, but scaled to have unit sum-of-squares. To get the correct scaling, one needs to multiply them by the square roots of the respective eigenvalues, so strictly speaking, $$\mathbf X_\mathrm{projected\:on\: axis \:\#i} \ne \mathbf K \mathbf E_i = \lambda_i \mathbf E_i,$$ but $$\mathbf X_\mathrm{projected\:on\: axis \:\#i} = \sqrt{\lambda_i} \mathbf E_i,$$ where $\mathbf E_i$ and $\lambda_i$ are the $i$-th eigenvector and eigenvalue of the kernel matrix.

This is easy to see by considering standard, non-kernel, PCA. There PCA amounts to an SVD decomposition of the data matrix: $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top,$$ where $\mathbf U \mathbf S$ are PCs and $\mathbf V$ are principal axes. The "kernel" matrix is the so called Gram matrix $$\mathbf K = \mathbf X \mathbf X^\top = \mathbf U \mathbf S^2 \mathbf U^\top,$$ and so one can see that its eigenvalues need to be square-rooted.

Turning now to your main question, you have a new data point $\mathbf x$ (a row vector) that needs to be projected on the principal axes. When faced with a question about kPCA, always think about how you would do it in standard PCA. You need to compute $\mathbf x \mathbf V$. But say you don't know $\mathbf V$ (that's the case in kPCA). Well, you can compute $\mathbf k = \mathbf x \mathbf X^\top$, which is a (row) vector of kernels between the new data point and all the old ones. And now $$\mathbf x \mathbf V = \mathbf x \mathbf V \mathbf S \mathbf U^\top \mathbf U \mathbf S^{-1} = \mathbf x \mathbf X^\top \mathbf U \mathbf S^{-1} = \mathbf k \mathbf U \mathbf S^{-1}.$$ Rewriting this in your kPCA notation, $$\mathbf x_\mathrm{projected} = \mathbf k \mathbf E \boldsymbol \Lambda^{-1/2},$$ where $\boldsymbol \Lambda$ is the diagonal matrix with eigenvalues $\lambda_i$ on the diagonal.

Let's consider the training dataset first. The data projected onto the principal axis (or axes) is principal component(s). In kPCA, eigenvectors of the kernel matrix directly give you principal components, but scaled to have unit sum-of-squares. To get the correct scaling, one needs to multiply them by the square roots of the respective eigenvalues, so $$\mathbf X_\mathrm{projected\:on\: axis \:\#i} = \sqrt{\lambda_i} \mathbf E_i,$$ where $\mathbf E_i$ and $\lambda_i$ are the $i$-th eigenvector and eigenvalue of the kernel matrix (and not $\mathbf K \mathbf E_i = \lambda_i \mathbf E_i$ as you wrote in your question).

This is easy to see by considering standard, non-kernel, PCA. There PCA amounts to an SVD decomposition of the data matrix: $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top,$$ where $\mathbf U \mathbf S$ are PCs and $\mathbf V$ are principal axes. The kernel matrix is the so called Gram matrix $$\mathbf K = \mathbf X \mathbf X^\top = \mathbf U \mathbf S^2 \mathbf U^\top,$$ and so one can see that to get $\mathbf{US}$ its eigenvalues need to be square-rooted.

Turning now to your main question, you have a new (test) data point $\mathbf x$ (a row vector) that needs to be projected on the principal axes. When faced with a question about kPCA, always think about how you would do it in standard PCA. You need to compute $\mathbf x \mathbf V$. But say you don't know $\mathbf V$ (that's the case in kPCA). Well, you can compute $\mathbf k = \mathbf x \mathbf X^\top$, which is a (row) vector of kernels between the new data point and all the old ones. And now $$\mathbf x \mathbf V = \mathbf x \mathbf V \mathbf S \mathbf U^\top \mathbf U \mathbf S^{-1} = \mathbf x \mathbf X^\top \mathbf U \mathbf S^{-1} = \mathbf k \mathbf U \mathbf S^{-1}.$$ Rewriting this in your kPCA notation, $$\mathbf x_\mathrm{projected} = \mathbf k \mathbf E \boldsymbol \Lambda^{-1/2},$$ where $\boldsymbol \Lambda$ is the diagonal matrix with eigenvalues $\lambda_i$ on the diagonal.