$\newcommand{\e}{\varepsilon}$Let $Y = X\beta + \e$ where $\e \sim \mathcal N(0, \Omega)$ with $\Omega$ known. Assume that $\Omega$ is not singular. Then there exists a matrix $L$ such that $L = \Omega^{-1/2}$ and we can multiply through our equation by $L$ to get
$$
LY = LX\beta + L\e
$$
and now $L\e \sim \mathcal N(0,I)$. Because we're left multiplying we're not changing $\beta$.
Since our errors are now iid we can happily apply the usual OLS procedure to get
$$
\hat \beta = \left((LX)^T(LX)\right)^{-1}(LX)^T(LY) = (X^T \Omega^{-1} X)^{-1}X^T \Omega^{-1} Y.
$$
This is generalized least squares. This answer covers exactly the point estimation of $\hat \beta$ but didn't seem to have the variance so I'll derive that next.
Because we are not estimating $\Omega$ we have
$$
E(\hat \beta) = (X^T \Omega^{-1} X)^{-1}X^T \Omega^{-1} X \beta = \beta
$$
and
$$
Var(\hat \beta) = \left((X^T \Omega^{-1} X)^{-1}X^T \Omega^{-1}\right)Var(Y)\left((X^T \Omega^{-1} X)^{-1}X^T \Omega^{-1}\right)
$$
$$
= (X^T \Omega^{-1} X)^{-1}
$$
so all together
$$
\hat \beta \sim \mathcal N\left(\beta, (X^T \Omega^{-1} X)^{-1} \right)
$$
since $\hat \beta$ is still a linear transformation of a Gaussian RV (i.e. $Y$).
Often GLS is done with $\Omega = \sigma^2 V$ where $V$ is known but $\sigma$ isn't, but here I've assumed the entire thing is known exactly.