II. (50 points) In an isosceles right triangle $\triangle ABC$, $CA = CB = 1$, and $P$ is any point on the boundary of $\triangle ABC$. Find the maximum value of $PA \cdot PB + PB \cdot PC + PC \cdot PA$.
(1) If $P \in AB$, as shown in Figure 7, let $AP = t$, then $PB = \sqrt{2} - t$, and $PC = \sqrt{t^2 - \sqrt{2} t + 1} (t \in [0, \sqrt{2}])$.
Thus, $m = PA \cdot PB + PB \cdot PC + PC \cdot PA$
$= PA \cdot PB + PC(PA + PB)$
$= t(\sqrt{2} - t) + \sqrt{t^2 - \sqrt{2} t + 1} \cdot \sqrt{2}$.
Let $u = \sqrt{t^2 - \sqrt{2} t + 1}$
$$
= \sqrt{\left(t - \frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}} \in \left[\frac{\sqrt{2}}{2}, 1\right].
$$

Then $t^2 - \sqrt{2} t = u^2 - 1$.
Thus, $m = 1 - u^2 + \sqrt{2} u = -\left(u - \frac{\sqrt{2}}{2}\right)^2 + \frac{3}{2} \leq \frac{3}{2}$.
The maximum value of $m$ is $\frac{3}{2}$, which occurs when $P$ is the midpoint of $AB$.
(2) If $P \in AC$, as shown in Figure 8, let $AP = t$, then
$$
\begin{array}{l}
PC = 1 - t, PB = \sqrt{t^2 - 2t + 2}. \\
\text{Thus, } m = PA \cdot PB + PB \cdot PC + PC \cdot PA \\
= \sqrt{t^2 - 2t + 2} + t(1 - t).
\end{array}
$$

We need to prove that for $t \in [0, 1]$,
$$
\sqrt{t^2 - 2t + 2} + t(1 - t) > 0 \\
\Leftrightarrow t^2(1 - t)^2 + t^2 + \left(t - \frac{1}{2}\right)^2 > 0.
$$

Inequality (2) is clearly true, so inequality (1) holds. By symmetry, $P \in BC$ is equivalent to $P \in AC$.
Therefore, the maximum value is $\frac{3}{2}$.