We first show that $Y$ lies on the circle $ABC$. To this end, let $Y'$ be the point on the ray $MD$, emanating from $M$, such that $MY' \cdot MD = MA^2$. The triangles $MAY'$ and $MDA$ are therefore similar and have opposite orientations. Since $MY' \cdot MD = MA^2 = MC^2$, so are the triangles $MCY'$ and $MDC$. Hence, using directed angles, $\angle AY'C = \angle AY'M + \angle MY'C = \angle MAD + \angle DCM = \angle CDA = \angle ABC$, so $Y'$ lies on the circle $ABC$.

Let the lines $AD$ and $BC$ meet at $Z$. The equality of directed angles $\angle PDZ = \angle PBC = \angle PBZ$ shows that $Z$ lies on the circle $BDP$; and the equality of directed angles $\angle Y'BZ = \angle Y'BC = \angle Y'AC = \angle Y'AM = \angle Y'DZ$ shows that so does $Y'$. Since the angle $ADC$ is acute, $MA \neq MD$, so $MY' \neq MD$, and $Y'$ is therefore the point where $MD$ crosses the circle $BDP$ again. Consequently, $Y' = Y$, so the latter lies on the circle $ABC$, by the preceding paragraph.

Next, we show that $FY$ is the internal bisectrix of the angle $AYC$. To this end, simply refer to the internal angle bisectrix theorem and the similar triangles above, to write
$$\frac{FA}{FC} = \frac{AD}{CD} = \frac{AD}{AM} \cdot \frac{CM}{CD} = \frac{YA}{YM} \cdot \frac{YM}{YC} = \frac{YA}{YC}.$$
Let now $BE$, the internal bisectrix of the angle $ABC$, cross the circle $ABC$ again at $B'$, so the latter is the midpoint of the arc $AC$ not containing $B$. The line $B'Y$ is therefore the external bisectrix of the angle $AYC$, so it is perpendicular to $FY$, by the preceding paragraph.

Let the line $\ell$ through $P$ and parallel to $AC$ meet the line $B'Y$ at $S$. To prove the required perpendicularity, we show $PQ$ and $PS$ perpendicular.

Use directed angles to write $\angle PSY = \angle(AC, B'Y) = \angle ACY + \angle CYB' = \angle ACY + \angle CAB' = \angle ACY + \angle B'CA = \angle B'CY = \angle B'BY = \angle PBY$, and infer that $S$ lies on the circle $BDP$. Restated, the line through $Y$ and perpendicular to $FY$ passes through the point where $\ell$ crosses the circle $BDP$ again, which is $S$. Similarly, the line through $X$ and perpendicular to $EX$ passes through the point where $\ell$ crosses the circle $BDP$ again, which is $S$. Consequently, $QX$ and $QY$ are perpendicular to $SX$ and $SY$, respectively, so $Q$ lies on the circle $BDP$ of which $QS$ is a diameter. The conclusion follows.

Begin by noticing that, since $\angle ABC = \angle ADC$, the circles $ABC$ and $ACD$ are reflections of one another across $M$.

We first show that $X$ lies on the circle $ACD$. To this end, let the ray $MB$, emanating from $M$, cross the circle $ACD$ at $X_1$, and let $X'$ be the reflection of $X_1$ across $M$. By the preceding, $X'$ lies on the circle $ABC$ and $AX_1CX'$ is a parallelogram, so, using directed angles, $\angle DX_1B = \angle DX_1A + \angle AX_1B = \angle DCA + \angle AX_1X' = \angle DCA + \angle CX'X_1 = \angle DCA + \angle CX'B = \angle DCA + \angle CAB = \angle(CD, AB)$. The equality of directed angles $\angle DPB = \angle PDC + \angle(CD, AB) + \angle ABP = \angle(CD, AB)$ therefore implies that $\angle DX_1B = \angle DPB$, so $X_1$ lies on the circle $BDP$. It follows that $X_1$ and $X$ coincide, so $X$ lies on the circle $ACD$. Similarly, $Y$ lies on the circle $ABC$.

Next, we show that $Q$ lies on the circle $BDP$. To this end, let the perpendicular bisectrix of the segment $AC$ cross the circle $ABC$ at $B'$ and $M_1$, and the circle $ACD$ at $D'$ and $M_2$, so that $B, D'$ and $M_1$ all lie in the same half-plane relative to the line $AC$. Notice that $B'$ and $D'$ lie on the bisectrices $BP$ and $DP$, respectively, of the angles $ABC$ and $ADC$, respectively. Further, notice that
$$\frac{BA \cdot X'A}{BC \cdot X'C} = \frac{\text{area } BAX'}{\text{area } BCX'} = \frac{MA}{MC} = 1,$$
and refer to the fact that $BE$ bisects the angle $ABC$, to write
$$\frac{EA}{EC} = \frac{BA}{BC} = \frac{X'C}{X'A} = \frac{XA}{XC},$$
and infer that $XE$ bisects the angle $AXC$, so $M_2$ lies on the line through $E, Q, X$.

Similarly, $M_1$ lies on the line through $F, Q, Y$. Use directed angles to write $\angle XQY = \angle M_2QM_1 = \angle QM_2M_1 + \angle M_2M_1Q = \angle XM_2D' + \angle B'M_1Y = \angle XDD' + \angle B'BY = \angle XDP + \angle PBY = \angle XBP + \angle PBY = \angle XBY$, and infer that $Q$ lies on the circle $BDP$.

Finally, since $M_1$ and $M_2$ are reflections of one another across $M$, the quadrangle $XM_1X'M_2$ is a parallelogram, so, using directed angles, $\angle XQP = \angle XBP = \angle X'BB' = \angle X'M_1B' = \angle XM_2M_1$. This shows that $PQ$ and $M_1M_2$ are parallel; since the latter is perpendicular to $AC$, so is the former.

If one of the vertices $B$, $D$ lies on the perpendicular bisectrix of the diagonal $AC$, so do both $P$ and $Q$, and the conclusion follows.

Assuming now that neither $B$ nor $D$ lie on the perpendicular bisectrix of the diagonal $AC$, we show that $P$ and $Q$ project orthogonally to the same point on the line $AC$. To this end, project $P$ and $Q$ on the line $AC$ to $P'$ and $Q'$, respectively, and notice that the two coincide if and only if, in terms of directed segments, $EP'/FP' = EQ'/FQ'$. Notice that
$$\frac{EP'}{FP'} = \frac{\tan \angle EFP}{\tan \angle FEP} = \frac{\tan \angle AFD}{\tan \angle AEB} \quad \text{and} \quad \frac{EQ'}{FQ'} = \frac{\tan \angle EFQ}{\tan \angle FEQ},$$
where angles are all directed, so it is sufficient to show that
$$(\tan \angle AEB)(\tan \angle EFQ) = (\tan \angle AFD)(\tan \angle FEQ).$$
To this end, let the line through $B$ and perpendicular to $BE$ meet the line $AC$ at $T$. We will prove that $B, E, T, X$ are concyclic, so, in terms of directed angles, $\angle FEQ = \angle TEX = \pi/2 + \angle EBM$. Similarly, $\angle EFQ = \pi/2 + \angle FDM$, so it is sufficient to show that
$$(\tan \angle AEB)(\tan \angle MBE) = (\tan \angle AFD)(\tan \angle MDF).$$
With reference to the sine law, straightforward calculations show that the left-hand member is $(\tan \angle EBA)^2$, the right-hand member is $(\tan \angle FDC)^2$, and the conclusion follows by equality of the two angles.

We still have to prove that $B, E, T, X$ are concyclic. Since $BT$ and $BX$ are the two bisectrices of the angle $ABC$, the cross-ratio $(T, E; A, C)$ is harmonic, so $ME \cdot MT = MA^2$.

It is therefore sufficient to show that $MA^2 = MB \cdot MX$. To this end, let the lines $AB$ and $CD$ meet at $W$, and let the lines $AD$ and $BC$ meet at $Z$. Use directed angles to write $\angle PDZ = \angle PBC = \angle PBZ$ and infer that $Z$ lies on the circle $BDP$; similarly, so does $W$, and the quadrangle $BDWZ$ is therefore cyclic.

Reflect $C$ in the circle $BDWZ$, centered at $O$, to obtain $C'$. Since $A$ lies on the polar of $C$ with respect to this circle, the lines $AC'$ and $OC$ are perpendicular, so $C'$ lies on the circle $\gamma$ on diameter $AC$. It follows that reflection in the circle $BDWZ$ sends $\gamma$ to itself, so the two circles are orthogonal. Consequently, the power of $M$ with respect to the circle $BDWZ$ is the square of the radius of $\gamma$; that is, $MB \cdot MX = MA^2$. This completes the proof.