8. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, let the symmetric points of vertex $A$ with respect to the plane $C_{1} B D$ and the line $B_{1} D$ be $P, Q$, respectively. Then the sine of the angle formed by the line $P Q$ and the plane $A_{1} B D$ is $\qquad$.
Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, since $A E \perp B_{1} D$, then $A D^{2}=D E \cdot D B_{1} \Rightarrow D E=\frac{1}{\sqrt{3}}$. $A F \perp$ plane $C_{1} B D$ and $A C$ intersects $B D$ at the midpoint of $A C$, then $A F \cong \frac{1}{3} A_{1} C$. Establish a right-handed coordinate system with $\overrightarrow{A B}, \overrightarrow{A D}, \overrightarrow{A A_{1}}$ as the positive directions of the $x, y, z$ axes, respectively. Then $D(0,1,0), B_{1}(1,0,1), A_{1}(0,0,1)$, $C(1,1,0)$, thus $\overrightarrow{E F}=\overrightarrow{E D}+\overrightarrow{D A}+\overrightarrow{A F}$ $=\frac{1}{3} \overrightarrow{B_{1} D}+\overrightarrow{D A}+\frac{1}{3} \overrightarrow{A_{1} C}=\left(0,-\frac{1}{3},-\frac{2}{3}\right)$.
Let the normal vector of plane $A_{1} B D$ be $\boldsymbol{n}=(x, y, z)$, then $\left\{\begin{array}{l}-x+y=0, \\ y-z=0\end{array} \Rightarrow \boldsymbol{n}=(1,1,1)\right.$. Therefore, $|\cos \langle\overrightarrow{E F}, \boldsymbol{n}\rangle|=\frac{1}{\sqrt{3} \cdot \sqrt{\frac{5}{9}}}=\frac{\sqrt{15}}{5}$, so the sine value of the angle formed by $P Q$ and plane $A_{1} B D$ is $\frac{\sqrt{15}}{5}$.