Let in triangle $ABC$, points $I_a$, $H$, and $D$ be the $A$-excenter, foot of the altitude from $A$, and foot of the angle bisector from $A$, respectively. Since $(AD, II_a) = -1$ and $\angle AHD = 90^\circ$, we have $\angle IHD = \angle DHI_a$. Further, $AIHQ$ is cyclic, So $\angle QAI = \angle IHD$, hence $\angle QAI = \angle DHI_a$ and $\triangle QAI \sim \triangle I_aHE$. We know $\angle AHE \sim \angle FAI$, therefore $\angle QIF \sim \angle I_aEA$ and we have
$$\angle FQI = \angle AI_aE = \angle IQB.$$
Now let $\omega$ be the circle centered at $M$, has the radius $MB = MC$, and $X$ and $Y$ be the intersections of line $FQ$ with $AB$ and $AC$, respectively. Point $I$ lies on the angle bisector of $\angle QYC$, since $I$ lies on the angle bisector of $\angle YQB$ and $\angle YCQ$. Therefore $Y$, $I$, and $O_c$ are collinear, where $O_c$ is the center of $\omega_c$. Now we have
$$\begin{align*} \angle IO_cC = 90^\circ - \frac{1}{2}\angle QYC &= \frac{1}{2}(\angle YQC + \angle YCQ) = \angle IQB + \angle ICB \\ &= \frac{1}{2}(\angle B - \angle C) + \frac{1}{2}\angle C = \frac{1}{2}\angle B \end{align*}$$
Therefore $O_c$ lies on $\omega$. Similarly we can show that $O_b$ lies on $\omega$, where $O_b$ is the center of $\omega_b$. Notice that $I$ is the $A$-excenter of triangle $AXY$ so
$$\angle O_b I O_c = \angle YIX = 90 - \frac{1}{2}\angle A = \angle B I_a C,$$
hence $BO_b = CO_c$. Finally
$$P_{\omega_b}^M = MO_b^2 - O_bB^2 = MO_c^2 - O_cC^2 = P_{\omega_c}^M,$$
we are done.