11.1.9 $\star \star P$ Inside or on the boundary of a square $A B C D$ with side length 1, find the maximum value of $P A \cdot P B \cdot P C \cdot P D$.
Let's assume $A C$ and $B D$ intersect at point $O$, and point $P$ is inside $\triangle A O D$, with $\angle P O D = \theta$ and $O P = x$. Using $x$ and $\theta$ (by the Law of Cosines), we can express $P A^{2}$, $P B^{2}$, $P C^{2}$, and $P D^{2}$ as follows:
$$
\begin{array}{c}
P A^{2}=O A^{2}+O P^{2}-2 O A \cdot O P \cos \angle A O P \\
=\frac{1}{2}+x^{2}-\sqrt{2} x \sin \theta \\
P C^{2}=\frac{1}{2}+x^{2}+\sqrt{2} x \sin \theta \\
P D^{2}=\frac{1}{2}+x^{2}-\sqrt{2} x \cos \theta, P B^{2}=\frac{1}{2}+x^{2}+\sqrt{2} x \cos \theta
\end{array}
$$

Thus, $P A^{2} \cdot P B^{2} \cdot P C^{2} \cdot P D^{2}=\left[\left(\frac{1}{2}+x^{2}\right)^{2}-2 x^{2} \sin ^{2} \theta\right]$
$$
\begin{aligned}
& {\left[\left(\frac{1}{2}+x^{2}\right)^{2}-2 x^{2} \cos ^{2} \theta\right] } \\
= & \left(\frac{1}{2}+x^{2}\right)^{4}+x^{4} \sin ^{2} 2 \theta-2 x^{2}\left(\frac{1}{2}+x^{2}\right)^{2},
\end{aligned}
$$

Clearly, the expression reaches its maximum when $\theta=45^{\circ}$. Adjusting $x$,
at this point, $P A \cdot P B \cdot P C \cdot P D=\left(\frac{1}{2}+x^{2}\right)^{2}-x^{2}=x^{4}+\frac{1}{4}$. When $x=\frac{1}{2}$ (i.e., $P$ is the midpoint of $A D$), it reaches the maximum value of $\frac{5}{16}$.