First solution. Let $\Gamma$ be a circle passing through $M, P, B$. Let $\Gamma$ touch the side $AC$ at $S$, and $T$ be the intersection of the lines $MS$ and $BC$. Since point $S$ lies on the side $AC$, it follows that $T$ lies on the extension of the side $BC$ (over $C$).

Since $AM = MB = MP$ and $MSPB$ is an inscribed quadrilateral, we see that $\angle MBP = \angle MPB = \angle MSB = \beta$, $\angle MBS = \angle ASM = \angle TSC = \varphi$. Let $\angle CAB = \alpha$. From the triangle $ASB$ we have $\alpha + \beta + 2\varphi = \pi$. Since $\angle SMB$ is an external angle of $\triangle ASM$, we have $\angle SMB = \alpha + \varphi$. Hence $\angle MTB = \pi - (\alpha + \varphi) - \beta = \varphi$. Thus, $\angle CTS = \angle CST(= \varphi)$, so $CS = CT$. By Menelaus' theorem ($\triangle ACB$ and points $M, S, T$),
$$\frac{AM}{MB} \cdot \frac{BT}{TC} \cdot \frac{CS}{SA} = 1 \iff BT = SA.$$
By the power of point theorem, $BT^2 = AS^2 = AM \cdot AB = BM \cdot BA$. It is equivalent to the fact that $BT$ is tangent to $\Gamma_1$, passing through $A, M, T$. Let $Q'$ be the intersection point of $\Gamma_1$ and the line $AC$. Since $\Gamma_1$ touches the line $BT$, we see that $Q'$ lies on the side $AC$ (not on the extension). Further
$$\begin{align*} \sim AM + \sim Q'T &= 2\angle TSC = 2\varphi = 2\angle STC = \\ &= [TC \text{ is tangent to } \Gamma_1] = \sim MQ'T \iff \\ \sim MA + \sim Q'T &= \sim MQ'T = \sim MQ' + \sim Q'T \iff \sim AM = \sim MQ'. \end{align*}$$

So $AM = MQ'$ (chords subtending the equal arcs). Therefore, $BM = AM = MQ'$, which gives that $AQ'B$ is a right-angled triangle ($\angle AQ'B = 90^\circ$). Thus $BQ' \perp AC$, i.e. points $Q$ and $Q'$ coincide, which completes the proof.

Second solution. Let $\Gamma$ be a circle passing through $M, P, B$. Let $\Gamma$ touch the side $AC$ at $S$. Since $APB$ and $AQB$ are right-angled triangles, we have $MA = MB = MQ = MP$, so $A, Q, P, B$ lie on the circle $\Gamma_0$ with the center at $M$. Consider the inversion (with $M$ as center) with respect to $\Gamma_0$. Since $\Gamma$ passes through the center of the inversion, its image is a line $\ell$, but $B$ and $P$ are fixed points, the line $\ell$ coincide with the line $BC$. Similarly, the image of the line $AC$ ($M \notin AC$) is a circle $\Gamma_1$ passing through $A, Q, M$, since $A$ and $Q$ are fixed points.
Therefore, $S$ as the tangency point of $\Gamma$ and the side $AC$ is transformed to $S'$ which is the tangency point of the line $BC$ and $\Gamma_1$. To complete the proof, it suffices to note that $S'$ lies on the ray $MS$, i.e. is the intersection point of the ray $MS$ and the line $BC$. Since $S$ lies on the side $AC$, we see that $S'$ lies on the extension of the side $BC$ (over $C$).