First, we can prove that $BP,\ CQ$ and $AP,\ AQ$ are isogonally conjugates in $\bigtriangleup ABC$, following a similar fashion as EGMO 2018, P5. Note that, $$LQ\cdot LR=LM\cdot LK=LA^2=LB^2$$so $$\angle ARB=\angle QAL+\angle LBQ=180^\circ-\angle A -\angle QAB-\angle QBA=180^\circ-\angle A-\angle PAC-\angle CBP=180^\circ -\angle APB$$The result follows.

Let $P'$ be the reflection of $P$ in line $KL$. Since $KM \cdot KL = KQ \cdot KP'$ and $KL$ bisects $\angle QKP'$, we obtain $$ \triangle KQL \sim \triangle KMP' $$i.e. A spiral similarity at $K$ takes $QL \to MP'$. It follows $$ QL,MP',\odot(KMQ),\odot(KP'L) $$concur. Hence $R \in MP', \odot(KP'L)$. Then PoP gives $$ MP' \cdot MR = MK \cdot ML = MA \cdot MB $$so points $A,B,P',R$ are concyclic, i.e. $R \in \odot(AP'B)$. But also $P \in \odot(AP'B)$ (isosceles trapezium), hence $$ \text{points } A,P,B,R \text{ are concyclic}$$which completes the proof.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.82380719967286cm); 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 = -12.898141875107362, xmax = 12.74947252423836, ymin = -5.417100780435011, ymax = 6.3176536682452085; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wqwqwq = rgb(0.3764705882352941,0.3764705882352941,0.3764705882352941); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wwqqcc = rgb(0.4,0.,0.8); pen ubqqys = rgb(0.29411764705882354,0.,0.5098039215686274); /* draw figures */ draw(circle((-2.333562187170986,2.072826043481851), 3.563826800548236), linewidth(2.) + wrwrwr); draw((-4.847037159176351,4.599346383674763)--(-4.945524112044927,-0.3517391479111737), linewidth(1.2) + wqwqwq); draw((-4.945524112044927,-0.3517391479111737)--(-0.16719963869743612,-0.7569680785130806), linewidth(1.2) + wqwqwq); draw((-0.16719963869743612,-0.7569680785130806)--(-4.847037159176351,4.599346383674763), linewidth(1.2) + wqwqwq); draw((-0.16719963869743612,-0.7569680785130806)--(-5.8966841091998745,2.1437036355899077), linewidth(1.2) + sexdts); draw(circle((-4.9791615449386315,-2.0427429049463037), 3.352050457075601), linewidth(1.6) + rvwvcq); draw((-5.8966841091998745,2.1437036355899077)--(-8.088739287079552,-0.791036656514778), linewidth(1.2) + linetype("4 4") + wvvxds); draw((-7.061822843287669,0.5838069581135685)--(1.2295597348579024,2.001948451373794), linewidth(1.6) + sexdts); draw((-8.088739287079552,-0.791036656514778)--(1.2295597348579024,2.001948451373794), linewidth(1.2) + sexdts); draw((-5.8966841091998745,2.1437036355899077)--(1.2295597348579024,2.001948451373794), linewidth(1.2) + dtsfsf); draw((-3.8878318420749,1.1266800309251617)--(-8.088739287079552,-0.791036656514778), linewidth(1.6) + wwqqcc); draw((-7.061822843287669,0.5838069581135685)--(-3.0707452528344503,0.7130128031693606), linewidth(1.2) + wwqqcc); draw(circle((-6.899382025892937,2.1636492943414676), 3.185080809481073), linewidth(1.2) + dotted + ubqqys); /* dots and labels */ dot((-4.847037159176351,4.599346383674763),dotstyle); label("$A$", (-5.114369499795724,4.679853407062501), NE * labelscalefactor); dot((-4.945524112044927,-0.3517391479111737),dotstyle); label("$B$", (-5.097484961020644,-0.8413907723884806), NE * labelscalefactor); dot((-0.16719963869743612,-0.7569680785130806),dotstyle); label("$C$", (0.052299365378595045,-1.044005237689434), NE * labelscalefactor); dot((-5.8966841091998745,2.1437036355899077),linewidth(4.pt) + dotstyle); label("$K$", (-6.262518136501128,2.265364362226139), NE * labelscalefactor); dot((1.2295597348579024,2.001948451373794),linewidth(4.pt) + dotstyle); label("$L$", (1.369293389834794,2.0289808193750263), NE * labelscalefactor); dot((-3.0707452528344503,0.7130128031693606),dotstyle); label("$Q$", (-3.0038021529107897,0.8808321826696237), NE * labelscalefactor); dot((-4.89628063561064,2.1238036178817947),linewidth(4.pt) + dotstyle); label("$M$", (-4.827332340619373,2.265364362226139), NE * labelscalefactor); dot((-3.8878318420749,1.1266800309251617),linewidth(4.pt) + dotstyle); label("$P$", (-3.8142600141146046,1.2691765744964512), NE * labelscalefactor); dot((-8.088739287079552,-0.791036656514778),linewidth(4.pt) + dotstyle); label("$R$", (-8.322431867060825,-0.5543536132121298), NE * labelscalefactor); dot((-7.061822843287669,0.5838069581135685),linewidth(4.pt) + dotstyle); label("$S$", (-7.258705924230817,0.7119867949188291), NE * labelscalefactor); dot((-4.925538518891552,0.6529663305280002),linewidth(4.pt) + dotstyle); label("$D$", (-4.844216879394453,0.30675786431692226), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $S$ be the second intersection of line $LP$ with the circumcircle of $\triangle PQR$. Let $D$ be the intersection of $PR$ with $QS$. First, we prove that $K,S,R$ are collinear: Consider the inversion at $L$ with radius $LA$. This takes the circumcircle of $ABC$ to the line $AB$, therefore $M \mapsto K$ and $K \mapsto M$. This in turn implies that it leaves the circumcircle of $\triangle KMQ$ fixed, hence $Q \mapsto R$. So, this means that it leaves the circumcircle of $\triangle PQR$ fixed, thus $P \mapsto S$. Finally, note that the circle $PQLM$ maps to a line, therefore $K,S,R$ are collinear. Now, we prove that $D$ lies on $AB$, and finish: We are reduced to a question purely about the cyclic quadrilateral $PQRS$ with circumcircle $\omega$, and the opposite sides meeting at $K,L$. Let $\omega_K$ be the circle with centre $K$ which is orthogonal to $\omega$ (this is the circle at $K$ with radius $KA$), and let $\omega_L$ be the circle with centre $L$ which is orthogonal to $\omega$ (this is the circle at $L$ with radius $LA$). Then the radical axis of $\omega_K$ and $\omega$ is the polar of $K$ wrt $\omega$, so the radical axis is $DL$ by brocard's theorem. Similarly the radical axis of $\omega_L$ and $\omega$ is $DK$. By radical axis theorem, $D$ lies on the radical axis of $\omega_K$ and $\omega_L$ which is $AB$. Thus, $DA \cdot DB = \mathrm{pow}_{\omega_K}(D) = \mathrm{pow}_{\omega}(D) = DP \cdot DR$, and therefore $A,P,B,R$ are concyclic.