Question 158, As shown in the figure, in $\triangle A B C$, $A B=B C=2, \angle A B C=120^{\circ}$, if point $P$ outside the plane $A B C$ and point $D$ on line segment $A C$ satisfy $P D=D A, P B=B A$, then the maximum volume of tetrahedron $P B C D$ is $\qquad$.

保留源文本的换行和格式，直接输出翻译结果如下：

Question 158, As shown in the figure, in $\triangle A B C$, $A B=B C=2, \angle A B C=120^{\circ}$, if point $P$ outside the plane $A B C$ and point $D$ on line segment $A C$ satisfy $P D=D A, P B=B A$, then the maximum volume of tetrahedron $P B C D$ is $\qquad$.
Question 158, Answer: The answer is $\frac{1}{2}$.
As shown in the figure, the condition is equivalent to $\triangle A B D \cong \triangle P B D$. Let $\mathrm{d}(\mathrm{A}, \mathrm{BD})=h$, then $\mathrm{d}(\mathrm{P}, \mathrm{ABC}) \leq \mathrm{d}(\mathrm{P}, \mathrm{BD})=$ $h$, equality holds if and only if plane $P B D \perp$ plane $A B C$.

Let $\mathrm{AD}=\mathrm{x}$, then $\mathrm{x} \in(0,2 \sqrt{3})$. According to the cosine theorem, we have: $\mathrm{BD}=\sqrt{2^{2}+\mathrm{x}^{2}-2 \times 2 \cdot x \cdot \cos 30^{\circ}}=$ $\sqrt{x^{2}-2 \sqrt{3} \cdot x+4}$, and thus $h=\frac{2 S_{\triangle A B D}}{B D}=\frac{2 \cdot x \cdot \sin 30^{\circ}}{\sqrt{x^{2}-2 \sqrt{3} \cdot x+4}}=\frac{x}{\sqrt{x^{2}-2 \sqrt{3} \cdot x+4}}$. Therefore,
$$
\begin{array}{l}
V_{P-B C D} \leq \frac{1}{3} \cdot S_{C B D} \cdot h=\frac{1}{3} \cdot \frac{2 \cdot(2 \sqrt{3}-x) \cdot \sin 30^{\circ}}{2} \cdot \frac{x}{\sqrt{x^{2}-2 \sqrt{3} \cdot x+4}} \\
=\frac{1}{6} \cdot \frac{x(2 \sqrt{3}-x)}{\sqrt{x^{2}-2 \sqrt{3} \cdot x+4}}
\end{array}
$$

Let $\mathrm{x}(2 \sqrt{3}-\mathrm{x})=\mathrm{t}$, then $\mathrm{t} \in(0,3)$, substituting into (1) we get:
$$
V_{P-B C D} \leq \frac{1}{6} \cdot \frac{t}{\sqrt{4-t}} \leq \frac{1}{6} \cdot \frac{3}{\sqrt{4-3}}=\frac{1}{2}
$$

Equality holds if and only if $D$ is the midpoint of $A C$, and $P D \perp$ plane $A B C$.