Let $N = CD \cap MP$. Clearly $\angle ABP = \angle MPB = 90^\circ - \angle CPN$. If $PM \perp CD$, then $\angle ABP = \angle ACD$. If $ABCD$ is cyclic, then $\angle PNC = 90^\circ$.

Let $N= (PM) \cap (CD)$ PM is median in right APM , so PM = MA=MB ABCD is cyclic iff <BAC = <BDC iff <MAP= <PDN iff < MPA = <PDN iff <CPN = < PDN iff 90^o - <NPD= <PND iff <NPD + < PDN = 90^o iff $PN \perp CD $ iff $PM \perp CD $

Let $MP \cap CD = X$. Note that $\angle MBP = \angle MPB = \angle XPD$. So $ABCD$ is cyclic iff $\angle XPD = \angle ACD = \angle PCD$. Now focus on $\Delta PCD$. There is a unique cevian $PX$ through $P$ such that $\angle XPD$ obtains a fixed value $\alpha$. But in the case that $\alpha = \angle PCD$, this cevian has to be the foot from $P$ onto $CD$, i. e. $ABCD$ is cyclic iff $PX \perp CD \iff PM \perp CD$. $\square$