It's easy to see that $AP=MQ$ and $PM=QC$. Note that $\angle APY = \angle ABC [\because P, Y, B, C$ concyclic$] = \angle CQX [\because Q, X, B, A$ concyclic$]$ Again, $\angle AYP = \angle BCA = \angle XCQ [\because P, Y, B, C$ concyclic$]$ $\therefore \triangle PYA \sim \triangle QCX$ $\therefore AP/XQ = YP/CQ \implies QM/XQ = YP/PM$ and we've $\angle XQM = \angle MPY (=\pi - \angle ABC)$ $\therefore \triangle XQM \sim \triangle MPY \implies \angle BXM = \angle XCQ + \angle XMQ = \angle AYP + \angle PYM = \angle AYM = \pi - \angle BYM \implies \angle BXM + \angle BYM = \pi$ $\therefore BXMY$ concyclic. $[QED]$

Suppose $(BXM)$ intersects $AC$ again at $M'$, where $M'\equiv M$ if $(BXM)$ is tangent to $AC$. We want to show that $AM'\cdot AM=AY\cdot AB$ or $AM'\cdot AM=AP\cdot AC$ or $AM'\cdot\dfrac{AC}{2}=AP\cdot AC$ or $AM'=2AP$. By Power of a Point, $CM'\cdot CM=CX\cdot CB=CQ\cdot CA$ or $CM'\cdot \dfrac{AC}{2}=CQ\cdot CA$ or $CM'=2CQ$. The rest is $AM'=AC-CM'=2(\dfrac{AC}{2}-CQ)=2AP$.

I used barycentric coordinates. Let $AP = MQ = u$ and $PM = QC = v$ where $u + v = \frac{b}{2}$. This implies that \begin{align*} P & = (b - u : 0 : u) \\ Q & = (v : 0 : b - v) \end{align*} We find the equations of the two circles. \begin{align*} (APQ) & : -a^2yz - b^2xz - c^2xy +(x+y+z)bvz = 0 \\ (BCP) & : -a^2yz - b^2xz - c^2xy +(x+y+z)bux = 0 \end{align*} From here we find $X$ and $Y$. \begin{align*} X & = (0:bv:a^2-bv) \\ Y & = (c^2 - bu : bu : 0) \end{align*} From here, we can find the equation of the circumcircle of $\triangle BXY$. $$\triangle BXY : -a^2yz - b^2xz - c^2xy + (x+y+z)(bux +bvz) = 0$$. We just need to check if $(1:0:1)$ satisfies the circle equation and it does. We are done.

By Power of a Point, $AY\cdot AB=AP\cdot AC$ and $CX\cdot CB=CQ\cdot CA$. Let $AC=b$ and $A=w$. By the condition, $CQ=\frac{b}{2}-w$. Thus, $AY\cdot AB=wb$ and $CX\cdot CB=\left(\frac{b}{2}-w\right)b$. Now, let the circumcircle of $BXY$ intersect $AC$ at $J$ and $K$. By Power of a Point, $AY\cdot AB=AJ\cdot AK$ and $CX\cdot CB=CJ\cdot CK$. Note that $CJ=b-AJ$ and $CK=b-AK$, so $CX\cdot CB=\left(b-AJ\right)\left(b-AK\right)$. Substituting in what we know, $AJ\cdot AK=wb$ and $\left(b-AJ\right)\left(b-AK\right)=\left(\frac{b}{2}-w\right)b$. There is a unique monic quadratic $P\left(x\right)$ with roots $AJ$ and $AK$. We would have $P\left(0\right)=wb$ and $P\left(b\right)=\left(\frac{b}{2}-w\right)b$. But notice $P\left(x\right)=\left(x-\frac{b}{2}\right)\left(x-2w\right)$ satisfies these same conditions, so this is $P\left(x\right)$. But $\frac{b}{2}$ is a root of $P\left(x\right)$, so either $AJ$ or $AK$ is $\frac{b}{2}$, meaning either $J$ or $K$ is $M$ and thus $M$ lies on the circumcircle of $BXY$ and $BXMY$ is cyclic. Q.E.D.

[asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.755, xmax = 24.513, ymin = -11.504, ymax = 10.6874; /* image dimensions */pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); /* draw figures */draw((-8.608,4.402)--(-0.952,-8.336), linewidth(2) + wrwrwr); draw((-0.952,-8.336)--(9.96,6.14), linewidth(2) + wrwrwr); draw(circle((4.719313527479312,-1.2603032061242232), 9.068036294911582), linewidth(2) + wrwrwr); draw(circle((-0.15727426166838726,0.8114258323651146), 9.181883704218178), linewidth(2) + wrwrwr); draw((-4.190241001697239,-2.9482369540726987)--(-1.8036725240608469,5.038897950946911), linewidth(2) + dtsfsf); draw((7.480327475939153,5.907897950946911)--(8.417767360743255,4.094054280986011), linewidth(2) + sexdts); draw((-4.190241001697239,-2.9482369540726987)--(0.676,5.271), linewidth(2) + dtsfsf); draw((0.676,5.271)--(8.417767360743255,4.094054280986011), linewidth(2) + sexdts); draw((-0.952,-8.336)--(0.676,5.271), linewidth(2) + wrwrwr); draw((-1.8036725240608469,5.038897950946911)--(0.676,5.271), linewidth(2) + dtsfsf); draw((-8.608,4.402)--(-1.8036725240608469,5.038897950946911), linewidth(2) + wrwrwr); draw((0.676,5.271)--(7.480327475939153,5.907897950946911), linewidth(2) + sexdts); draw((7.480327475939153,5.907897950946911)--(9.96,6.14), linewidth(2) + wrwrwr); draw(circle((3.535229052857004,-1.971980921441259), 7.786909746816359), linewidth(1.6) + linetype("2 2") + rvwvcq); /* dots and labels */dot((-8.608,4.402),dotstyle); label("$A$", (-8.52,4.6374), NE * labelscalefactor); dot((-0.952,-8.336),dotstyle); label("$B$", (-0.8486,-8.0918), NE * labelscalefactor); dot((9.96,6.14),dotstyle); label("$C$", (10.0656,6.3798), NE * labelscalefactor); dot((0.676,5.271),linewidth(4pt) + dotstyle); label("$M$", (0.7728,5.4602), NE * labelscalefactor); dot((-1.8036725240608469,5.038897950946911),dotstyle); label("$P$", (-1.6956,5.2908), NE * labelscalefactor); dot((7.480327475939153,5.907897950946911),linewidth(4pt) + dotstyle); label("$Q$", (7.573,6.0894), NE * labelscalefactor); dot((-4.190241001697239,-2.9482369540726987),linewidth(4pt) + dotstyle); label("$Y$", (-4.0914,-2.7436), NE * labelscalefactor); dot((8.417767360743255,4.094054280986011),linewidth(4pt) + dotstyle); label("$X$", (8.5168,4.2986), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let's first prove that triangles $MQX$ and $YPM$ are similar. Our given circles show $BYPC$ cyclic and $BXQA$ cyclic, so $$\angle ABC=180-\angle YPM=\angle APY$$and $$\angle ABC=180-\angle XQM=\angle CQX$$ These show that by AA similarity, we have $$\triangle ABC\sim \triangle APY \sim \triangle XQC$$ Now, this implies $$\frac{BC}{AB}=\frac{PY}{AP}=\frac{QC}{XQ}$$ Since $PQ=AM=CM$, we have $AP=MQ$ and $CQ=MP$. This means from the above equation, we have $$\frac{PY}{MP}=\frac{QM}{XQ}$$ We showed above that $\angle YPM=\angle MQX$, so by SAS similarity, we have $\triangle YPM\sim \triangle MQX$ This means that $\angle XMQ=\angle MYP$, so we have $$\angle XMY=180^{\circ}-\angle YMP-\angle XMQ=180^{\circ}-\angle YMP-\angle MYP=\angle YPM=\angle YPC=180^{\circ}-\angle YBX$$ so $BXMY$ is cyclic, as desired.

The idea is to use the so called "lin pop trix". Let $\omega, \Omega$ denote the circles $\odot(BAQ)$ and $\odot(BXY)$ respectively. Define a function $f : \mathbb{R}^2 \mapsto \mathbb{R}$ as: \[f(P) = \text{Pow}(P, \Omega)-\text{Pow}(P, \omega)\]Since $C$ lies on the radical axis of $\Omega, \omega$ we have $f(C) = 0$. On the other hand $f(A) = AP \cdot AC$ since $P \in \omega$. Since $f(M)$ is the average $f(A)$ and $f(C)$ due to the aforementioned tricx, we obtain \[f(M) = AP \cdot (AC/2) = MQ \cdot MA\]since $AP = MQ$ (from $PQ = AC/2$) and $MA = AC/2$. On the other hand, \[f(M) = \text{Pow}(M, \Omega) - \text{Pow}(M, \omega) = \text{Pow}(M, \Omega) - (-MQ \cdot MA)\]implying $\text{Pow}(M, \Omega) = 0$, that is, $M \in \Omega$. Hence $BXMY$ is cyclic, as desired. $\square$