As shown in the figure, since plane $P B D$ is perpendicular to plane $A B C$, draw $P E \perp$ $B D$ at $E$, and connect $E C$. Then $P E \perp$ plane $A B C$ and $P E=E C$, obviously $E C \leqslant B C$, equality holds when $B D \perp B C$. Therefore, the maximum distance from point $P$ to plane $P B C$ is 2.
5. As shown in the figure, in $\triangle A B C$, $\angle A B C=120^{\circ}, A B=A C=2$. Take a point $D$ on side $A C$ (excluding $A, C$), and fold $\triangle A B D$ along segment $B D$ to get $\triangle P B D$. When plane $P B D$ is perpendicular to plane $A B C$, the maximum distance from point $P$ to plane $P B C$ is $\qquad$