4. As shown in the figure, given the ellipse $\frac{x^{2}}{2} + y^{2}=1, P A \perp A B, C B \perp$ $A B$, and $D A=3 \sqrt{2}, C B$ $=\sqrt{2}$. Point $P$ moves on the arc $\overparen{A B}$. Then the minimum value of the area of $\triangle P C D$ is $\qquad$
4. $4-\sqrt{6}$.

The equation of the line $CD$ is: $x+y-2 \sqrt{2}=0$.
Let the parametric coordinates of the moving point $P$ on the ellipse be $P(\sqrt{2} \cos \theta, \sin \theta), \theta \in [0,2 \pi)$.
The distance from $P$ to $CD$ is
$$
\begin{aligned}
d & =\frac{|\sqrt{2} \cos \theta+\sin \theta-2 \sqrt{2}|}{\sqrt{2}} \\
& =\frac{|\sqrt{3} \sin (\theta+\alpha)-2 \sqrt{2}|}{\sqrt{2}} \\
& =\frac{2 \sqrt{2}-\sqrt{3} \sin (\theta+\alpha)}{\sqrt{2}} .
\end{aligned}
$$

where $\alpha=\arcsin \frac{\sqrt{6}}{3}$, when $\sin (\theta+\alpha)=1$, i.e., $\theta=\frac{\pi}{2}-\arcsin \frac{\sqrt{6}}{3}$, $d$ reaches its minimum value $2-\frac{\sqrt{6}}{2}$. Therefore, the minimum value of the area of $\triangle PCD$ is $4-\sqrt{6}$.