Cesko-Slovacko-Poljsko 2013

Let $D$ be the midpoint of arc $BC$, $N$ the midpoint of $BC$, $X$ the orthogonal projection of $P$ onto $AC$. Then points $P,X,K,L,N,C$ are concyclic (Fig. 7).

Fig. 7

We have $\angle PNX=\angle PCA=\angle PDA$ therefore $XN\parallel KL$. It follows that $LN=KX$, and $\angle LCN=\angle KCA$. Obviously $\angle BAK=\angle LAC$, so $K,L$ are isogonal conjugates with respect to triangle $ABC$. Thus following equalities hold: $\angle MBC=\angle CBL=\angle KBA$ and $\angle BCM=\angle LCB=\angle ACK$. This implies that $A,M$ are isogonal conjugates with respect to triangle $KBC$. Using this we get $\angle BNM=\angle LNB=\angle LKC=\angle BKM$ therefore points $B,M,N,K$ are concyclic.

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Alternative solution.

In the first solution, it is proven that points $K,L$ are isogonal conjugates with respect to $ABC$. Thus angle bisectors of $\angle CBA$ and $\angle BCA$ are angle bisectors of $\angle LBK$ and $\angle KCL$, respectively. From angle bisector theorem we deduce that $BK/BL=KC/LC$. This means that points $K,L$ lie on Apollonian circle $\omega$ of points $B,C$. Since $\omega$ is self-symmetric with respect to line $BC$, we get that $M$ lies on $\omega$ (Fig. 8).

Fig. 8

Let $\omega$ intersect segment $BC$ at point $Y$. Then $KY$ is angle bisector of both angles $\angle MKL$ (because $Y$ is the midpoint of arc $LM$) and $\angle BKC$ (because $K$ lies on Apollonian circle). Therefore $\angle BKM=\angle LKC$. We end proof as we did in the first solution.