8. As shown in the figure, given a regular tetrahedron $ABCD$ with edge length 1, $M$ is the midpoint of $AC$, and $P$ lies on the line segment $DM$. Then the minimum value of $AP + BP$ is $\qquad$
8. Analysis: $\sqrt{1+\frac{\sqrt{6}}{3}}$. Let $\angle B D M=\theta$. In $\triangle B D M$, $B D=1, B M=M D=\frac{\sqrt{3}}{2}$. Therefore, $\cos \theta=\frac{\sqrt{3}}{3}, \sin \theta=\frac{\sqrt{6}}{3}$. As shown in the figure, rotate $\triangle B D M$ around $D M$ so that $\triangle B D M$ lies in the plane $A C D$, at this point $B$ is at $B^{\prime}$.
Connect $A B^{\prime}, B^{\prime} P$. The minimum value sought is the length of $A B^{\prime}$. It is easy to see that
$\angle A D B^{\prime}=\theta+30^{\circ}$.
Thus, $A B^{\prime 2}=A D^{2}+D B^{\prime 2}-2 A D \cdot D B^{\prime} \cos$ $\angle A D B^{\prime}$
$$
\begin{array}{l}
=1^{2}+1^{2}-2 \cos \left(\theta+30^{\circ}\right) \\
=2-2\left(\cos \theta \cdot \cos 30^{\circ}-\sin \theta \cdot \sin 30^{\circ}\right)=1+\frac{\sqrt{6}}{3} .
\end{array}
$$

Therefore, $A B^{\prime}=\sqrt{1+\frac{\sqrt{6}}{3}}$.