Bulgarian National Olympiad - Final Round

Let $D$ be the reflection of $P$ across $C$, so $AB=BD$ and $O^{\prime }$ be the circumcenter of $△ABD$. As $△A{O}^{\prime }B\cong △B{O}^{\prime }D$ and $MB=CD$, we have $\angle O^{\prime }CB=\angle O^{\prime }MA$, so $MBC{O}^{\prime }$ is cyclic. Now $\angle O^{\prime }CM=\angle O^{\prime }BM=\angle AOM$, hence $C{O}^{\prime }\parallel AP$. Therefore $O^{\prime }$ must be the midpoint of $AD$, implying that $\angle ABC=90^{\circ }$. $\square$