3. As shown in the figure, in the quadrilateral pyramid $P-ABCD$, the base $ABCD$ is a rectangle, $PA \perp$ plane $ABCD$, $PA=AB=2$, $AD=6$, points $M, N$ are on edges $AB, BC$ respectively. When the perimeter of the spatial quadrilateral $PMND$ is minimized, the cosine value of the dihedral angle $P-MN-D$ is . $\qquad$
Extend $DA$ to point $P'$, such that $AP' = AP = 2$, and point $D'$ is symmetric to point $D$ with respect to point $C$.

Then $PM + MN + ND$
$$
= P'M + MN + ND' \geqslant P'D',
$$

with equality holding when $P', M, N, D'$ are collinear. Thus, $\triangle P'AM \sim \triangle P'DD'$
$$
\Rightarrow \frac{P'A}{AM} = \frac{P'D}{DD'} = 2 \Rightarrow AM = 1.
$$

Construct $AE \perp AM$ at point $E$, and connect $PE$. Then $\angle PEA$ is the plane angle of the dihedral angle $P-MN-D$. In $\triangle PAE$, $AE = \frac{2}{\sqrt{5}} \Rightarrow PE = \sqrt{4 + \frac{4}{5}} = \frac{2\sqrt{6}}{\sqrt{5}} \Rightarrow \cos \angle PEA = \frac{\sqrt{6}}{6}$, so the cosine value of the dihedral angle $P-MN-D$ is $\frac{\sqrt{6}}{6}$.