Maybe the reason this "isn't obvious" to you is that it's not exactly true.

If $n$ is large and $p$ is not too far from $1/2,$ then $X\sim\mathsf{Binom}(n, p)$ $X$ is approximately $\mathsf{Norm}(np, \sqrt{np(1-p)}.$ and $\hat p = X/n$ is approximately $\mathsf{Norm}(p, \sqrt{p(1-p)/n}).$ This is follows from the Central Limit Theorem and other considerations.

However, that does not quite answer your question. In your expression, notice that you have $\hat p$ instead of $p.$ Again, if $n$ is sufficiently large, the Law of Large Numbers says (rougly) that $\hat p \approx p.$

Therefore, it is not exactly correct to say that $\frac{\hat p - p}{\sqrt{\hat p(1-\hat p)/n}}$ has a standard normal distribution. [In case $n$ is in the thousands, this expression is nearly standard normal.]

Hypothesis tests. Usually, the test statistic for testing $H_0: p = p_0$ against $H_a: p \ne p_0$ would be $z = \frac{\hat p - p_0}{p_0(1-p_0)/n},$ where $p_0$ is the value of $p$ specified in the null hypothesis.

Confidence intervals. However, if you are making a confidence interval, there is no specified hypothetical value $p = p_0.$ The Wald 95% confidence interval is of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p )}{n}}.$ Strictly speaking, this is an asymptotic confidence interval. That is, it is approximately correct only if $n$ is very large.

A slight modification of the Wald CI is the Agresti-Coull CI, which has been shown to be more accurate than the Wald interval for small and moderate $n.$ Let $\check p = \frac{X+2}{n+4}.$ Then the A-C 95% CI is of the form $\check p \pm 1.96\sqrt{\frac{\check p(1-\check p )}{n+4}}.$

Maybe the reason this "isn't obvious" to you is that it's not exactly true.

If $n$ is large and $p$ is not too far from $1/2,$ then $X\sim\mathsf{Binom}(n, p)$ $X$ is approximately $\mathsf{Norm}(np, \sqrt{np(1-p)}.$ and $\hat p = X/n$ is approximately $\mathsf{Norm}(p, \sqrt{p(1-p)/n}).$ This is follows from the Central Limit Theorem and other considerations.

However, that does not quite answer your question. In your expression, notice that you have $\hat p$ instead of $p.$ Again, if $n$ is sufficiently large, the Law of Large Numbers says (rougly) that $\hat p \approx p.$

Therefore, it is not exactly correct to say that $\frac{\hat p - p}{\sqrt{\hat p(1-\hat p)/n}}$ has a standard normal distribution. [In case $n$ is in the thousands, this expression is nearly standard normal.]

Hypothesis tests. Usually, the test statistic for testing $H_0: p = p_0$ against $H_a: p \ne p_0$ would be $z = \frac{\hat p - p_0}{p_0(1-p_0)/n},$ where $p_0$ is the value of $p$ specified in the null hypothesis.

Confidence intervals. However, if you are making a confidence interval, there is no specified hypothetical value $p = p_0.$ The Wald 95% confidence interval is of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p )}{n}}.$ Strictly speaking, this is an asymptotic confidence interval. That is, it is approximately correct only if $n$ is very large.

A slight modification of the Wald CI is the Agresti-Coull CI, which has been shown to be more accurate than the Wald interval for small and moderate $n.$ Let $\check p = \frac{X+2}{n+4}.$ Then the A-C 95% CI is of the form $\check p \pm 1.96\sqrt{\frac{\check p(1-\check p )}{n+4}}.$

Note_ See @Glen_b's link to Slutsky's Theorem as justification for use of $\hat p$ when $n$ is very large.

If $n$ is large and $p$ is not too far from $1/2,$ then $X\sim\mathsf{Binom}(n, p)$ $X$ is approximately $\mathsf{Norm}(np, \sqrt{np(1-p)}.$ and $\hat p = X/n$ is approximately $\mathsf{Norm}(p, \sqrt{p(1-p)/n}).$ This is due to the Central Limit Theorem.

However, that does not quite answer your question. In your expression, notice that you have $\hat p$ instead of $p.$ Again, if $n$ is sufficiently large, the Law of Large Numbers says (rougly) that $\hat p \approx p.$

Therefore, it is not exactly correct to say that $\frac{\hat p - p}{\sqrt{\hat p(1-\hat p)/n}}$ has a standard normal distribution. [In case $n$ is in the thousands, this expression is nearly standard normal.]

Hypothesis tests. Usually, the test statistic for testing $H_0: p = p_0$ against $H_a: p \ne p_0$ would be $z = \frac{\hat p - p_0}{p_0(1-p_0)/n},$ where $p_0$ is the value of $p$ specified in the null hypothesis.

Confidence intervals. However, if you are making a confidence interval, there is no specified hypothetical value $p = p_0.$ The Wald 95% confidence interval is of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p )}{n}}.$ Strictly speaking, this is an asymptotic confidence interval. That is, it is approximately correct only if $n$ is very large.

A slight modification of the Wald CI is the Agresti-Coull CI, which has been shown to be more accurate than the Wald interval for small and moderate $n.$ Let $\check p = \frac{X+2}{n+4}.$ Then the A-C 95% CI is of the form $\check p \pm 1.96\sqrt{\frac{\check p(1-\check p )}{n+4}}.$

Maybe the reason this "isn't obvious" to you is that it's not exactly true.

If $n$ is large and $p$ is not too far from $1/2,$ then $X\sim\mathsf{Binom}(n, p)$ $X$ is approximately $\mathsf{Norm}(np, \sqrt{np(1-p)}.$ and $\hat p = X/n$ is approximately $\mathsf{Norm}(p, \sqrt{p(1-p)/n}).$ This is follows from the Central Limit Theorem and other considerations.

However, that does not quite answer your question. In your expression, notice that you have $\hat p$ instead of $p.$ Again, if $n$ is sufficiently large, the Law of Large Numbers says (rougly) that $\hat p \approx p.$

Therefore, it is not exactly correct to say that $\frac{\hat p - p}{\sqrt{\hat p(1-\hat p)/n}}$ has a standard normal distribution. [In case $n$ is in the thousands, this expression is nearly standard normal.]

Hypothesis tests. Usually, the test statistic for testing $H_0: p = p_0$ against $H_a: p \ne p_0$ would be $z = \frac{\hat p - p_0}{p_0(1-p_0)/n},$ where $p_0$ is the value of $p$ specified in the null hypothesis.

Confidence intervals. However, if you are making a confidence interval, there is no specified hypothetical value $p = p_0.$ The Wald 95% confidence interval is of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p )}{n}}.$ Strictly speaking, this is an asymptotic confidence interval. That is, it is approximately correct only if $n$ is very large.

A slight modification of the Wald CI is the Agresti-Coull CI, which has been shown to be more accurate than the Wald interval for small and moderate $n.$ Let $\check p = \frac{X+2}{n+4}.$ Then the A-C 95% CI is of the form $\check p \pm 1.96\sqrt{\frac{\check p(1-\check p )}{n+4}}.$