5. Given that the base $A B C$ of the regular tetrahedron $P-A B C$ is an equilateral triangle, and the center $O$ of the circumscribed sphere of the regular tetrahedron satisfies $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=0$. Then the cosine value of the dihedral angle $A-P B-C$ is $(\quad)$.
(A) $\frac{1}{6}$
(B) $\frac{\sqrt{2}}{8}$
(C) $\frac{1}{5}$
(D) $\frac{\sqrt{3}}{3}$
5. C.

As shown in Figure 1.
From $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=0$, we know that $O$ is the circumcenter of $\triangle A B C$.
Let the side length of $\triangle A B C$ be $a$.
Then the height of this triangular pyramid $P O=O B=\frac{\sqrt{3}}{3} a$,
the length of the lateral edges $P A=P B=P C=\frac{\sqrt{6}}{3} a$,
and the slant height of the side face $P D=\sqrt{P B^{2}-\left(\frac{B C}{2}\right)^{2}}=\sqrt{\frac{5}{12}} a$.
Draw $C E \perp P B$ at point $E$, and connect $A E$.
It is easy to prove that $\angle A E C$ is the plane angle of the dihedral angle $A-P B-C$.
In $\triangle P B C$, by $P B \cdot C E=P D \cdot B C$, we know $C E=\sqrt{\frac{5}{8}}$.
Thus, $\cos \angle A E C=\frac{A E^{2}+C E^{2}-A C^{2}}{2 A E \cdot C E}=\frac{1}{5}$.