$\angle YXC=\angle YDC=\angle BAC$,so $XY\parallel AB$. Then $\frac{CK}{DL}=\frac{AX}{BY}=\frac{AP}{BP}=\frac{DP}{CP}$. So $CM^2=CK\cdot CP=DL\cdot DP=DM^2$,so $M$ is the midpoint of $CD$. Let $CPDN$ be a parallelagram,so $M$ is the midpoint of $PN$,too. Then $WM\parallel LN$ and $ZM\parallel KN$. Notice that $\angle LDN=\angle KCN$ and $\frac{CK}{LD}=\frac{DP}{CP}=\frac{CN}{DN}$,so $\bigtriangleup LDN\backsim \bigtriangleup KCN$. Finally,$\angle MZP=\angle NKP=180^\circ -\angle NKC=180^\circ -\angle NLD=\angle NLP=\angle MWP$.$\Box$.

Attachments:

Lemma: $M$ is the midpoint of $CD$. Proof: Let $O$ be the center of $(PKL)$. Then: $AP\cdot CP = BP\cdot DP\implies (PX+AX)\cdot CP = (PY+BY)\cdot DP\implies AX\cdot CP = BY \cdot DP\implies\implies CK\cdot CP = DL\cdot DP\implies Pow_{(PKL)}C = Pow_{(PKL)}D\implies\boxed{CO=DO}$ That means the perpendicular through $M$ is the perpendicular bissector of $CD$ and we are done.$_{\blacksquare}$ Now notice that $\angle OZC = \angle OMC = 90^{\circ}\implies \angle MZC = \angle MOC$, analogously: $\angle MWD = \angle MOD\implies \angle MZC = \angle MWD\implies \angle MZP = \angle MWP.$ $_{\blacksquare}$