Let $\alpha=\angle BAP$. At least one of the triangles $BPA$, $PMA$ has an angle $\ge\frac\pi2$ at $P$ (and a nonzero angle at $B$ or $M$, respectively). From the sum of angles in a triangle, we conclude $\alpha<\frac\pi2$. Construct $X$ such that $\angle PBX=\frac\pi2-\alpha$ and $XP\perp BC$. Then $\angle BXP=\alpha$ and $A,X$ are on the same side of $BC$. Therefore by the inverse of the inscribed angle theorem, $A$ is on the circumscribed circle of $BPX$. Reflecting the figure at $BP$, we find $\angle PXM=\alpha$, hence $A$ is also on the circumscribed circle of $PMX$. These two circles intersect exactly in $P$ and $X$. From $A\ne P$ we conclude $A=X$ and from this $|AB|=|AM|$
Translate the figure by $\overrightarrow{BM}$. Then $B\mapsto M$, $M\mapsto C$, $A\mapsto A''$. Since $|A''M|=|A''C|$, let $c$ be the circle araound $A''$ through $M$ and $C$. Because $\angle MA''C=2\alpha$, the circle $c$ is the locus of all points $X$ with $\angle MXC=\alpha$. Hence $A\in c$. We conclude that $|BM|=|AA''|=|A''M|=|AB|$, i.e. $BMA$ is equilateral, hence $\alpha=\frac\pi6$. The rest is easy: $\angle BAC=3\alpha=\frac\pi2$, $\angle CBA=\frac\pi2-\alpha=\frac\pi3$, $\angle BCA=\frac\pi2-2\alpha=\frac\pi6$.