Solution with bash : Let $Q=AB\cap CD$, we define $CD=2a$, $QD=b$, $QA=x$, $AB=y$ and $PQ=z$ $\Longrightarrow$ $MD=MC=a$. Since $ABCD$ is cyclic $\Longrightarrow$ $QA.QB=QD.QC$ we get $$x(x+y)=b(b+2a)$$By Stewart theorem in $\triangle PQB$ with cevian $PA$ we get: $$z^2y+a^2x=a^2(x+y)+xy(x+y)$$$$\Longrightarrow z^2=a^2+x(x+y)=a^2+2ab+b^2$$Hence $QP=z=a+b=QM$ $\Longrightarrow$ $AB, CD$ and the perpendicular bisector of $MP$ are concurrent in $Q.$

Let $Q=AB\cap CD$. Note that $QA\cdot QB=QC\cdot QD$, so the power of $Q$ to the circle centered at $P$ with radius $PA=PB$ is equal to the power of $Q$ to the circle centered at $M$ with radius $MC=MD$. Since these circles are congruent and $Q$ lies on their radical axis, $Q$ lies on the perpendicular bisector of their centers, as desired.

This problem was proposed by Burii.

Pythagorean theorem works. Let $N$ be the midpoint of $AB$ and $AB$ intersects $CD$ at $Q$. Notice that \[QP^2 - QM^2 = QN^2 + NP^2 - QM^2 = (OQ^2 - ON^2) - (OQ^2 - OM^2) + NP^2 = OM^2 - ON^2 + NP^2 = (OM^2 + PB^2) - (ON^2 + NB^2) = (OM^2 + MC^2) - OB^2 = OC^2 - OB^2 = 0.\] So, QP = QM and Q lies in perp bisector of MP

Let $O$ be the center of $(ABCD)$. Let $E=AB\cap CD$ and $F$ be the midpoint of $AB$. We have $EP^2 - CM^2 = EP^2 - PA^2 = EF^2 - FA^2$ and $EM^2 - CM^2 = EM^2 - MD^2 = EO^2 - OC^2 = EO^2 - OA^2$. So it suffices to show $EF^2 - FA^2 = EO^2 - OA^2$ but that is true because $EO^2 - EF^2 = FO^2 = OA^2 - FA^2$. $\square$