5. In the tetrahedron $P-ABC$, $\triangle ABC$ is an equilateral triangle with side length $2 \sqrt{3}$, $PB=PC=\sqrt{5}$, and the dihedral angle $P-BC-A$ is $45^{\circ}$. Then the surface area of the circumscribed sphere of the tetrahedron $P-$ $ABC$ is $\qquad$
5. $25 \pi$.

As shown in Figure 2, take the midpoint $D$ of $BC$, and connect $AD$, $PD$.
From the problem, we know that
$AD \perp BC, PD \perp BC, PD=\sqrt{2}$.
Thus, $\angle PDA$ is the plane angle of the dihedral angle $P-BC-A$, $\angle PDA=45^{\circ}, BC \perp$ plane $PAD$, plane $PAD \perp$ plane $ABC$.
Construct $PO_1 \perp AD$ at point $O_1$, then $PO_1 \perp$ plane $ABC$.
Therefore, $PO_1=O_1D=1, O_1A=2, O_1$ is the circumcenter of $\triangle ABC$, and the tetrahedron $P-ABC$ is a regular tetrahedron.

Let the center of the circumscribed sphere of the tetrahedron $P-ABC$ be $O$, and the radius be $R$.
Then point $O$ lies on the line $PO_1$, and
$\left|PO_1-PO\right|^{2}+O_1A^{2}=OA^{2}$.
Thus, $(R-1)^{2}+2^{2}=R^{2} \Rightarrow R=\frac{5}{2}$.
Therefore, the surface area of the circumscribed sphere of the tetrahedron $P-ABC$ is $4 \pi R^{2}=25 \pi$.