9. (15 points) In the tetrahedron $S-ABC$, it is known that $SC \perp$ plane $ABC$, $AB=BC=CA=4\sqrt{2}$, $SC=2$, and $D$, $E$ are the midpoints of $AB$, $BC$ respectively. If point $P$ moves on $SE$, find the minimum value of the area of $\triangle PCD$.

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9. As shown in Figure 3, in the tetrahedron $S-ABC$, the base $ABC$ is an equilateral triangle, and $D$ is the midpoint of $AB$. Then
$$
\begin{array}{l}
|CD|=\frac{\sqrt{3}}{2}|AB| \\
=\frac{\sqrt{3}}{2} \times 4 \sqrt{2} \\
=2 \sqrt{6}.
\end{array}
$$

Thus, to find the minimum value of the area of $\triangle PCD$, we only need to find the minimum distance from point $P$ to line $CD$.

In the plane $SBC$, draw $PF \perp BC$ through point $P$, with the foot of the perpendicular at $F$.
Since $SC \perp$ plane $ABC$, we have
plane $ABC \perp$ plane $SBC$, with the intersection line being $BC$.
By the property of perpendicular planes, we know
$PF \perp$ plane $ABC$.
Draw $FG \perp CD$ in plane $ABC$ through point $F$, with the foot of the perpendicular at $G$, and connect $PG$.
Then, by the theorem of three perpendiculars, we know $CD \perp PG$.
Thus, the distance from point $P$ to $CD$ is $PG$.
Let $PF = x$. Given $SC = 2$, then $x \in (0, 2)$.
It is easy to see that Rt $\triangle SCE \sim \mathrm{Rt} \triangle PFE$.
Thus, $\frac{FE}{CE} = \frac{PF}{SC}$
$\Rightarrow FE = \frac{PF}{SC} \cdot CE = \frac{x}{2} \cdot 2 \sqrt{2} = \sqrt{2} x$.
Therefore, $CF = CE - FE = 2 \sqrt{2} - \sqrt{2} x$.
In Rt $\triangle FGC$,
$FG = CF \sin \angle FCG$
$= (2 \sqrt{2} - \sqrt{2} x) \sin 30^{\circ} = \sqrt{2} - \frac{\sqrt{2}}{2} x$.
In Rt $\triangle FGP$,
$PG^2 = PF^2 + FG^2 = x^2 + \left(\sqrt{2} - \frac{\sqrt{2}}{2} x\right)^2$
$= \frac{3}{2} x^2 - 2 x + 2 = \frac{3}{2} \left(x - \frac{2}{3}\right)^2 + \frac{4}{3} \geqslant \frac{4}{3}$.
Thus, when $x = \frac{2}{3}$, $PG$ reaches its minimum value $\frac{2 \sqrt{3}}{3}$.
Therefore, the minimum value of the area of $\triangle PCD$ is
$\frac{1}{2}|CD|(PG)_{\text{min}} = 2 \sqrt{2}$.