As shown in the figure, let the midpoint of $BC$ be $D$, and connect $AD, PD$. Draw $PE \perp AD$ at $E$. Then $PB=\sqrt{5}$, $BD=\sqrt{3} \Rightarrow PD=\sqrt{2}$, and $\angle PDE=45^{\circ}$ $\Rightarrow PE=DE=1$, which means $E$ is the centroid of $\triangle ABC$. Thus, the circumcenter $O$ of the sphere lies on the line $PE$, and we have $(R-1)^{2}+4=R^{2} \Rightarrow -2R+5=0 \Rightarrow R=\frac{5}{2}$. Therefore, the surface area of the circumscribed sphere of the tetrahedron $P-ABC$ is $S=25\pi$.
5. In the triangular pyramid $P-ABC$, $\triangle ABC$ 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 triangular pyramid $P-ABC$ is