II. As shown in Figure 3, a wooden stick $(AB)$ of length $2a$ is leaning against a wall $(ON)$ perpendicular to the ground $(OM)$, with an inclination angle $(\angle ABO)$ of $60^{\circ}$. If the end $A$ of the stick slides down along the line $ON$, and the end $B$ slides to the right along the line $OM$ $(NO \perp OM)$, the midpoint $P$ of the stick will also move. Given that when the end $A$ slides down to $A'$, $AA' = (\sqrt{3}-\sqrt{2})a$. Find the length of the path that the midpoint $P$ has traveled to $P'$.
II. $\because P O=\frac{1}{2} A B=a$ is a constant, $O$ is a fixed point,
$\therefore$ the path from $P$ to $P^{\prime}$ is an arc with $O$ as the center and $a$ as the radius. Connecting $O P, O P^{\prime}$, it is easy to see that $O P=P A=O P^{\prime}=P^{\prime} B^{\prime}=a$. In the right triangle $\triangle A B O$, $O A=A B \sin 60^{\circ}=2 a \cdot \frac{\sqrt{3}}{2}=\sqrt{3} a$, then $O A^{\prime}=O A-A A^{\prime}=\sqrt{3} a-(\sqrt{3}-\sqrt{2}) a=\sqrt{2} a$.

In the right triangle $\triangle A^{\prime} B^{\prime} O$, $A^{\prime} B^{\prime}=2 a$,
$$
\text { then } \begin{aligned}
O B^{\prime} & =\sqrt{A^{\prime} B^{\prime 2}-O A^{\prime 2}}=\sqrt{(2 a)^{2}-(\sqrt{2} a)^{2}} \\
& =\sqrt{2} a .
\end{aligned}
$$

Thus $\angle A^{\prime} B^{\prime} O=45^{\circ}, \angle B^{\prime} O P^{\prime}=\angle A^{\prime} B^{\prime} O=45^{\circ}$.
But $\angle A O P=\angle O A B=30^{\circ}$,
$\therefore \angle P O P^{\prime}=90^{\circ}-30^{\circ}-45^{\circ}=15^{\circ}$.
$\therefore$ the length of $\overparen{P P}$ is $\frac{15^{\circ}}{360^{\circ}} \cdot 2 \pi a=\frac{1}{12} \pi a$.