Let the circle through $A,K,M$ cut $AC$ in $T$. It is well known ($\triangle BMK \cong \triangle CTK$) that $CT=BM=AM$ and $CP=AP$ and $\angle PCT=\angle BAP$ thus we have $\triangle AMP \cong\triangle PTC$ thus $PT=PM$ so $AMPT$ is cyclic which means $AMKPT$ is cyclic too.

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Let point $K$ be the midpoint of arc $BAC$ $A,M,P,K$ are concyclic $\Leftrightarrow AP\cdot sin\angle MAK=AM\cdot sin\angle PAK+AK\cdot sin\angle MAP$ $(\angle MAK=90^o+\frac{A}{2}\Rightarrow sin\angle MAK=cos\angle \frac{A}{2};\angle PAK=90^o;\angle MAP=\frac{A}{2}$ $AM=\frac{c}{2};AK=2R\cdot sin\angle ANK=2R\cdot sin\angle B-(90^o-\frac{A}{2})=-2R\cdot cos(B+\frac{A}{2}))$ $\Leftrightarrow AP\cdot cos\frac{A}{2}= \frac{c}{2}-2R\cdot cos(B+\frac{A}{2})\cdot sin\frac{A}{2}$ $\Leftrightarrow \frac{c-b}{2}=2R\cdot cos(B+\frac{A}{2})\cdot sin\frac{A}{2}$ $\Leftrightarrow sinC-sinB=2cos(B+\frac{A}{2})\cdot sin\frac{A}{2}$ $\Leftrightarrow (2)cos\frac{B+C}{2}sin\frac{C-B}{2}=(2)cos(B+\frac{\pi -(B+C)}{2}sin\frac{A}{2}$ Obviously $cos\frac{B+C}{2}=sin\frac{A}{2}$ $\Leftrightarrow sin\frac{C-B}{2}=cos(B+\frac{\pi -(B+C)}{2})=-sin\frac{-(C-B)}{2}=sin\frac{C-B}{2}$ So we are done.

I have a nice solution by complex numbers! Let the circumcircle of $\triangle ABC$ is a unit circle and let $A=x^2$, $B=y^2$, $C=z^2$. Then we have that P is an intersection of lines ($xz$; $-xz$) and ($x^2$; $-yz$). So, we may use a formula and get that $P=\frac{z(x^2-yz)}{z-y}$. And it's easy to check that points $A=x^2$, $K=yz$, $M=\frac{x^2+y^2}{2}$, $P=\frac{z(x^2-yz)}{z-y}$ are concyclic since $\frac{(A-M)(K-P)}{(A-P)(K-M)} \in \mathbb{R}$.

valsidalv007 wrote: I have a nice solution by complex numbers! Let the circumcircle of $\triangle ABC$ is a unit circle and let $A=x^2$, $B=y^2$, $C=z^2$. Then we have that P is an intersection of lines ($xz$; $-xz$) and ($x^2$; $-yz$). So, we may use a formula and get that $P=\frac{z(x^2-yz)}{z-y}$. And it's easy to check that points $A=x^2$, $K=yz$, $M=\frac{x^2+y^2}{2}$, $P=\frac{z(x^2-yz)}{z-y}$ are concyclic since $\frac{(A-M)(K-P)}{(A-P)(K-M)} \in \mathbb{R}$. Incredibly beautiful! Really nice solution. Actually I wanted to do the same, but found a synthetic solution

Perform an inversion at Centre $A$ with ratio $\sqrt{AB.AC}$ followed by an reflection over $A$ angle bisector.It takes $M$ to the reflection of $A$ over $C$.Call this point $C'$ As $AK$ is the external angle bisecter of $\angle A$ $K$ goes to the point where $AK$ meets $BC$. Call this point $J$. The perpendicular bisector of $AC$ passes through $O$,the circumcentre of $ABC$ and the midpoint of $AC$.As after inversion $O$ goes to the reflection of $A$ over $BC$ (call this point $O'$ )and midpoint of $AC$ goes to reflection of $A$ over $B$ (call this point $B'$) so the perpendicular bisector bocomes the circumcircle of $AO'B'$.As $P$ is the intersection of perpendicular bisecter of $AC$ and $\angle A$-internal bisector so P goes to the intersection of $(AB'O')$ and $\angle A$-internal bisector .Call this point $L$. Note that $\angle ALB'=90^\circ$.So $BL'||AK$. So the inverted problem becomes: In triangle $AB'C'$ $B,C$ are midpoints of $AB'$ and $AC'$.External angle bisector of angle $A$ cuts $BC$ at $J$.The foot of the perpendicular from $B'$ to $\angle A$-internal bisector is $L$.Prove that $C',L,J$ are collinear. proof:Let internal $\angle A$ bisector cut $BC$ at $X$.We will use phantom point method to probe the (inverted) problem. Let $JC \cap AX=L'$ We will show that $\angle AL'B'= 90^{\circ}$. Let $AJ\cap BL= R$ and $BL\cap AC=G$.Note that $-1=(J,X;B,C)=A(J,X;B,C)=(R,L';B,G)$ Now taking perspective from $C'$ $-1=C(R,L;B,G)=(CR \cap BC,P_{\infty};J,C)$.Call $CR \cap BC= N$ .From the last result we get $C$ is the midpoint of $JN$.So $C'N||AJ$.So As $C',N,R$ are collinear and $R$ lies also on $AJ$ so $R \equiv P_ {\infty}$ across line $AJ$.So $B'L'||AJ$ implies $\angle B'L'A= 90^{\circ}$.We are done.$\blacksquare$