22. Solution: (1) From the given conditions, the polar coordinate equations of the circles containing arcs $AB$, $BC$, $CD$ are $\rho = 2\cos\theta$, $\rho = 2\sin\theta$, $\rho = -2\cos\theta$ respectively.
Thus the polar coordinate equation of $M_1$ is $\rho = 2\cos\theta \left(0 \leq \theta \leq \frac{\pi}{4}\right)$, the polar coordinate equation of $M_2$ is $\rho = 2\sin\theta \left(\frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}\right)$,
the polar coordinate equation of $M_3$ is $\rho = -2\cos\theta \left(\frac{3\pi}{4} \leq \theta \leq \pi\right)$.
(2) Let $P(\rho, \theta)$. From the given conditions and (1), we have
If $0 \leq \theta \leq \frac{\pi}{4}$, then $2\cos\theta = \sqrt{3}$, solving gives $\theta = \frac{\pi}{6}$;
If $\frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}$, then $2\sin\theta = \sqrt{3}$, solving gives $\theta = \frac{\pi}{3}$ or $\theta = \frac{2\pi}{3}$;
If $\frac{3\pi}{4} \leq \theta \leq \pi$, then $-2\cos\theta = \sqrt{3}$, solving gives $\theta = \frac{5\pi}{6}$.
In summary, the polar coordinates of $P$ are $\left(\sqrt{3}, \frac{\pi}{6}\right)$ or $\left(\sqrt{3}, \frac{\pi}{3}\right)$ or $\left(\sqrt{3}, \frac{2\pi}{3}\right)$ or $\left(\sqrt{3}, \frac{5\pi}{6}\right)$.