8. 14 .

Let $P_{1}$ and $P_{2}$ be the symmetric points of $P$ with respect to the planes $\alpha$ and $\beta$. Then the distance $P_{1} P_{2}$ is the minimum value of the perimeter of $\triangle P A B$, which is
$$
P_{1} P_{2}=\sqrt{6^{2}+10^{2}-2 \times 6 \times 10 \cos 120^{\circ}}=14 .
$$
8. Given the dihedral angle $\alpha-l-\beta$ is $60^{\circ}$, a point $P$ inside this dihedral angle is 3 units away from plane $\alpha$ and 5 units away from plane $\beta$. If $A \in \alpha, B \in \beta$, then the minimum perimeter of $\triangle P A B$ is $\qquad$