As shown in Fig. 1, draw the circumcircles of $\triangle ABC$, $\triangle ABP$ and $\triangle ACP$, respectively. Let the extension of $AP$ meet $\odot O$ at $D$, join $BD$, and draw the line tangent to $\odot O$ at $A$, intersecting $\odot O_1$ and $\odot O_2$ at $E$ and $F$ respectively.

It is clear that $\triangle AMC \sim \triangle ABD$, hence
$$\frac{AB}{BD} = \frac{AM}{MC}.$$
Since $\triangle EAB \sim \triangle PDB$, we have $\frac{AB}{BD} = \frac{AE}{PD}$.
Consequently, $\frac{AM}{MC} = \frac{AE}{PD}$, i.e.
$$AE = \frac{AM \times PD}{MC},$$
and, similarly,
$$AF = \frac{AM \times PD}{MB}.$$

Fig. 1

It follows that
$$AE = AF. \qquad ①$$

Draw the perpendicular lines $O_1E' \perp AE$ with foot $E'$, and $O_2F' \perp AF$ with foot $F'$. Since $E'$, $F'$ are the midpoints of $AE$, $AF$ respectively, it follows from ① that $A$ is the midpoint of $E'F'$.
In the right-angled trapezoid $O_1E'F'O_2$, $AO$ is the extension of the median, and hence it bisects the segment $O_1O_2$.

Solution 2:

As shown in Fig. 2, draw segments $AO_1$, $OO_1$, $AO_2$, $OO_2$. Denote by $Q$ the intersection of $AO$ and $O_1O_2$. Then

Fig. 2
$$\frac{O_1Q}{QO_2} = \frac{S_{\triangle OO_1}}{S_{\triangle OO_2}} = \frac{AB \times OO_1}{AC \times OO_2},$$
where $\frac{AB}{AC} = \frac{\sin\angle ACB}{\sin\angle ABC}$.
Since $\angle OO_1Q = \angle BAP = \angle CAM$, and $\angle OO_2Q = \angle CAP = \angle BAM$, it follows that
$$\begin{aligned} \frac{OO_1}{OO_2} &= \frac{OO_1}{OQ} \times \frac{OQ}{OO_2} \\ &= \frac{\sin\angle OQO_1}{\sin\angle OO_1Q} \times \frac{\sin\angle OO_2Q}{\sin\angle OQO_2} \\ &= \frac{\sin\angle OO_2Q}{\sin\angle OO_1Q} = \frac{\sin\angle BAM}{\sin\angle CAM}, \end{aligned}$$
and thus
$$\frac{O_1Q}{QO_2} = \frac{\sin\angle ACM}{\sin\angle CAM} \times \frac{\sin\angle BAM}{\sin\angle ABM} = \frac{AM}{CM} \times \frac{BM}{AM} = \frac{BM}{CM}.$$
Note that $M$ is the midpoint of $BC$, and therefore $O_1Q = QO_2$, i.e. line $AO$ bisects segment $O_1O_2$.