4. The five vertices of a regular quadrilateral pyramid $P-ABCD$ lie on the same sphere. If the side length of the base of the regular quadrilateral pyramid is 4, and the lateral edge length is $2 \sqrt{6}$, then the surface area of the sphere is ( ).
(A) $36 \pi$
(B) $12 \pi$
(C) $24 \pi$
(D) $48 \pi$
4. A.

As shown in Figure 3, let $O$ be the center of the circumscribed sphere, and $P H \perp$ the base $A B C D$ at $H$. Then
$$
\begin{array}{c}
O P=O C=R, \\
H C=2 \sqrt{2}, \\
O H=\sqrt{R^{2}-8}, \\
P H=R+\sqrt{R^{2}-8} . \\
P H^{2}+H C^{2}=P C^{2}, \text { thus, } \\
\left(R+\sqrt{R^{2}-8}\right)^{2}+(2 \sqrt{2})^{2}=(2 \sqrt{6})^{2} .
\end{array}
$$

Solving for $R$ gives $R=3$.
Therefore, $S_{\text {sphere }}=4 \pi R^{2}=36 \pi$.