$\textbf{Lemma:}$ Let $ABC$ be a triangle, the tangents by $B$ and $C$ meet at $P$, let $M$ be a point in the arc $AC$. If $PM\cap \odot (ABC)=X$, $AM\cap BC=K$ and $J$ is the midpoint of $BC$ $\Longrightarrow$ $AJKX$ is cyclic $\textbf{Proof:}$ It's well-know $PX$ is the $X$-symmedian of $\triangle BXC$ $\Longrightarrow$ $\measuredangle BXJ=\measuredangle XMC=\measuredangle MAC=\alpha$, on the other hand since $ABCX$ is cyclic $\Longrightarrow$ $\measuredangle CBX=\measuredangle CAX=\beta$. Finally $\measuredangle XAM$ $=$ $\measuredangle MAC$ $+$ $\measuredangle CAX$ $=$ $\alpha+\beta$ $=$ $\measuredangle JXB$ $+$ $\measuredangle JBX$ $=$ $\measuredangle XJK$ hence $\measuredangle KAX=\measuredangle KJX$ then $AJKX$ is cyclic $\textbf{In the Problem:}$ Let $PM\cap \odot (ABC)=X$ then by lemma $AKJX$ is cyclic. Since $AP$ is the $A$-symmedian of $\triangle ABC$ $\Longrightarrow$ $\measuredangle CAX$ $=$ $\measuredangle BAJ=\alpha$, since $PX$ is the $X$-symmedian of $\triangle BXC$ $\Longrightarrow$ $\measuredangle CXM$ $=$ $\measuredangle CAM$ $=$ $\measuredangle BXJ$ $=$ $\beta$. Let $\measuredangle PAJ=\theta$ $\Longrightarrow$ since $ABCX$ is cyclic we get $\measuredangle BAC$ $+$ $\measuredangle BXC$ $=$ $180^{\circ}$ $\Longrightarrow$ $\measuredangle JXM$ $=$ $180^{\circ}-2\alpha-2\beta-\theta$, but $\measuredangle QAK$ $=$ $\measuredangle XAK$ $=$ $\alpha+\beta$ $\Longrightarrow$ $\measuredangle JAQ$ $=$ $\measuredangle JAQ$ $=$ $2\alpha+2\beta+\theta$ $\Longrightarrow$ $\measuredangle JAQ$ $=$ $\measuredangle JXQ=180^{\circ}$ $\Longrightarrow$ $JXAQ$ is cyclic $\Longrightarrow$ $Q\in \odot(AJXK)$. Let $\measuredangle XAC$ $=$ $\measuredangle XBC$ $=$ $\omega$ , since $\measuredangle LAK$ $=$ $90^{\circ}$ $\Longrightarrow$ $\measuredangle LAX$ $=$ $\measuredangle LAM$ $-$ $\measuredangle XAM$ $=$ $90^{\circ}-\omega-\beta$, on the other hand $\measuredangle XJC$ $=$ $\measuredangle XBJ$ $+$ $\measuredangle JXB$ $=$ $\omega+\beta$ $\Longrightarrow$ $\measuredangle LJX$ $=$ $\measuredangle LJC$ $+$ $\measuredangle CJX$ $=$ $90+\beta+\omega$ $\Longrightarrow$ $\measuredangle XJL$ $+$ $\measuredangle LAX$ $=$ $180^{\circ}$ hence $LAJX$ is cyclic $\Longrightarrow$ $L\in \odot (AJKXQ)$ Finally $A,L,Q,K,J$ are concyclic.

Beautiful problem, i will sketch my solution, 1-$LJ\perp BK$ and $LA\perp AK$, then $L,A,J,K$ belong to the circumference with diameter $LK$ 2- Denote $\angle MAP=\alpha$. Consider $S$ the intersection of $AP$ and $\Gamma$, $T$ point on $PM$ such that $\angle STP=\angle PAQ=2\alpha$ and $U$ on $PK$ such that $OU\perp PK$. 3- $K$ is the image of $U$ after inversion with center on $P$ and radius $PB$, then $B,P,U,C$ lie on a circle and by power of point, $A,P,U,M$ lie on a circle, which implies $\angle MUK=\alpha$ (1) 4-using the inversion $A,S,U,K$ lie on a circle, which implies $\angle SUP=\alpha$ (2) 5- $\angle MOS=2\alpha $, then (1),(2) imply $O,S,U,M$ lie on a circle which also contains $T$. 6-Finally our magic inversion send $(O,S,T,U)$ to $(J,A,Q,K)$, this fact and the first step imply that $L,A,J,K,Q$ lie on a circle.

Let us assume without loss of generality, $AB>AC$. Let $A'$ be a point on the circumcircle of triangle $ABC$ such that $AA' \parallel BC$ and let $F$ be the other intersection point of the line $AP$ with the circumcircle of triangle $ABC$. Let $Q'$ be the point symmetric to the point $Q$ in the line $AM$. Denote by $\angle B,\angle C$, the respective measures of angles $ABC$ and $ACB$. The key observation is that for any point $X$ on the circumcircle of triangle $ABC$, the line $XP$ is a symmedian in triangle $XBC$. We shall use this fact later on. Lemma 1. The points $L,J,A,K$ are concyclic. Proof: Indeed, the circle with diameter $LK$ contains the point $J$. As the lines $PR$ and $AK$ are perpendicular and $AL \parallel PR$, we see that $\angle LAK=90^{\circ}$ and the claim holds. Lemma 2. The points $F,J,Q',M,K$ are concyclic. Proof: Observe that the points $A',J,F$ are collinear which follows since $AF$ is a symmedian in triangle $ABC$; and so, the lines $AJ$ and $FJ$ are symmetric in the line $BC$. We obtain $$\angle FMK=\angle FBA=\angle AJC=\angle FJK$$yielding that the points $F,J,M,K$ are concyclic. We see that the lines $MQ'$ and $MP$ are symmetric in the lines $AK$ and the lines $MP$ and $MJ$ are symmetric in the bisector of angle $BMC$. These two observations lead to the fact that $\angle JMQ'=\angle B-\angle C$. Note that $$\angle JMQ'=\left(\angle B-\angle C\right)=\angle AFA'=\angle Q'FJ.$$This forces the conclusion that the points $F,J,Q',M,K$ are concyclic. In order to finish the proof of the requested assertion, simply note that $$\angle AQK=\angle AQ'K=180^{\circ}-\angle FQ'K=180^{\circ}-\angle FJK=\angle AJB$$whence, the points $A,Q,K,J$ are concyclic. Together with Lemma 1., the conclusion follows.

Solution. Define $X=PQ\cap \Gamma; Y=AP\cap \Gamma$. A straightforward application of Brockard's theorem to $AMXY$ leads us to conclude that $X,Y,K$ are collinear (taking into account that $BC$ is the polar of $P$ wrt $\Gamma$). As in the previous posts, we can prove that $LAJK$ is cyclic. Clearly $\angle YAK=\angle QAK$, then $\angle QAK=\angle YAK=\angle YAM=\angle MXK=\angle QXK$, then $QAXK$ is cyclic. Similarly, we can prove using Brockard's theorem that $Z=YM\cap AX, B,C$ are collinear. It's well-known that $(B,C;Z,K)$ is a harmonic bundle, and also it's well-known that in this case $KC\cdot KB=KZ\cdot KJ$. Then $KM\cdot KA= KX\cdot KY=KB\cdot KC=KZ\cdot KJ\cdot$. Consider a inversion $\Phi$ with center $K$ and radius $r=\sqrt{KM\cdot KA}$. Then $\Phi(M)=A,\ \Phi(N)=J,\ \Phi(Y)=X$; since $Y,Z,M$ are collinear, this implies that the image of line $YM$ under $\Phi$ is the circle which passes through $A,J,X$ and the center of inversion $K$, i.e. points $A,J,X,K$ are cyclic. Finally, $LAJK, QAXK, AJXK$ are cyclic quadrilaterals, therefore points $L,A,J,X,K, Q$ are cyclic. $\blacksquare$

Another proof: As in the previous posts we prove that $J,K,A$ and $L$ are concyclic. As @above said, we can prove that the points he named $X,Y$ and $K$ are collinear. @above provided a proof of this, but it can be also proved by assuming that $AM$ and $XY$ meet line $BC$ in different points $K_1$ and $K_2$ and then prove that $\frac{K_1C}{K_1B}=\frac{K_2C}{K_2B}$ by using the lenght ratios obtained by the harmonic quadrilaterals. Now simple angle chasing gives us that $\angle KYA=\angle XYC+\angle CYA=\angle CMX+\angle CMK=\angle AMQ$, so we obtain $\triangle AYK\sim AMQ$. Now from this similarity we also obtain that $\triangle AYM\sim \triangle AKQ$ , so $\angle AYM=\angle AKQ$ and now again by simple angle chasing we get that $\angle AJK + \angle KQA =180^{\circ} $, so $JKQA$ is cyclic and we're done. $\blacksquare $

Solved with Rg230403 Let $AP \cap \Gamma = E$, $ME \cap BC = F$ and $AF \cap \Gamma = D$ Since $(A,E;B,C) = -1$, projecting through $M$, we get that $(B,C;F,K) = -1$. Then, projecting through $D$, we see that $(M,D;B,C) = -1 \implies M,D,P$ are collinear. Also, projecting through $E$, we see that $E,D,K$ are collinear as well. Now, since $\angle ADK = 180 - \angle EDA = 180 - \angle AJB = \angle AJK$ and so $AJDK$ is cyclic. Since $LA || PR$ and $PR \perp AK$, $\angle LAK = \angle LJK = 90^\circ$ and so $L$ also lies on $(AJDK)$ Also, $\angle AQD = \angle AMP - \angle MAQ = \angle AMD - \angle MAP = 180 - \angle AEK - \angle EAK = \angle AKE = \angle AKD$ and so $Q$ also lies on $(AJDK)$. So, the points $L,J,A,Q,K$ all lie on a circle, as desired. $\blacksquare$

My solution: We have that $RP \perp AM$ and $AL \parallel PR$ so $\angle LAK =90$ and $\angle PJK=90$ so $ALKJ$ is cyclic. Let $U= PM \cap \odot (ABC)$ and $V= PA \cap \odot (ABC)$ by Brockard's theorem we have that $U,V,K$ are collinear and with the symmedian property we have that: $$\angle AJB= 180 - \angle JAB - \angle ABJ = 180- \angle UAC - \angle AUC = \angle ACU = \angle AVU $$This implies that $AKVJ$ is cyclic, finally $\angle QAK= \angle PAK = \angle MVK =\angle QVK$ this implies that $AQLV$ is cyclic. So we have that $L,J,A,Q,K$ and $V$ all lie on a circle, as desired. $\blacksquare$