8. As shown in the figure, in the right triangle $\triangle ABC$, the right-angle side $AC=2\sqrt{2}$, and $D$ is on $AB$ such that $AD=1$. A moving point $P$ in the plane satisfies that the area of $\triangle PAB$ is half the area of $\triangle ABC$. Then the maximum value of $|PC-PD|$ is
$3$