8. Let $P$ be a moving point on the edge $AB$ of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$. Then the minimum value of the dihedral angle formed by the plane $P D B_{1}$ and the plane $A D D_{1} A_{1}$ is $\qquad$ .
8. $\arctan \frac{\sqrt{2}}{2}$.

As shown in Figure 2, extend $B_{1} P$ to intersect $A_{1} A$ at point $Q$. Then $Q$ is a common point of plane $P D B_{1}$ and plane $A D D_{1} A_{1}$. Therefore, $D Q$ is the edge of the dihedral angle.

Since $P A \perp$ plane $A D D_{1} A_{1}$, draw $A H \perp Q D$ at point $H$ within plane $A D D_{1} A_{1}$, and connect $P H$. Then $\angle P H A$ is the plane angle of the dihedral angle.
Let the edge length of the cube be $1, P A=x$. Then
$$
\begin{array}{l}
A Q=\frac{x}{1-x}, D Q=\frac{\sqrt{1-2 x+2 x^{2}}}{1-x}, \\
A H=\frac{x}{\sqrt{1-2 x+2 x^{2}}} .
\end{array}
$$

Thus, $\tan \angle P H A=\frac{A P}{A H}=\sqrt{2 x^{2}-2 x+1}$
$$
=\sqrt{2\left(x-\frac{1}{2}\right)^{2}+\frac{1}{2}} \geqslant \frac{\sqrt{2}}{2} \text {. }
$$

The equality holds if and only if $P$ is the midpoint of $A B$.