5. As shown in Figure 2, in $\triangle A B C$, $\angle A B C=120^{\circ}$,
$A B=B C=2$. Take a point $D$ on side
$A C$ (excluding $A$ and $C$), and fold $\triangle A B D$
along line segment $B D$, to get
$\triangle P B D$. When plane
$P B D \perp$ plane $A B C$, the maximum distance from point $P$ to plane $A B C$ is $\qquad$
5. 2 .

From the problem, we know that $\angle B A D=\angle B C A=30^{\circ}$.
By the cosine rule, we get $A C=2 \sqrt{3}$.
Let $A D=x$. Then
$$
0<x<2 \sqrt{3}, D C=2 \sqrt{3}-x \text {. }
$$

In $\triangle A B D$, by the cosine rule we get
$$
B D=\sqrt{x^{2}-2 \sqrt{3} x+4} .
$$

In $\triangle P B D$, $P D=A D=x, P B=B A=2$,
$$
\angle B P D=30^{\circ} \text {. }
$$

Let the distance from point $P$ to plane $A B C$ be $d$. Then
$$
\begin{array}{l}
S_{\triangle P B D}=\frac{1}{2} B D \cdot d=\frac{1}{2} P D \cdot P B \sin \angle B P D \\
\Rightarrow d=\frac{x}{\sqrt{x^{2}-2 \sqrt{3} x+4}}=\frac{1}{\sqrt{1-\frac{2 \sqrt{3}}{x}+\frac{4}{x^{2}}}} .
\end{array}
$$

From $0<x<2 \sqrt{3}$, we get $d_{\max }=2$.