Since $PM \perp AC$ $\Rightarrow$ $ADMP$ is cyclic. Claim: $K$ lies on $ADMP$. Proof: Since $PC$ is tangent to $\Omega$ $\Rightarrow \angle KCP = \angle CAK$ and as $\angle PAC = \angle PCA \Rightarrow \angle PAK = \angle KCA = \angle KCS = \angle KDS = \angle KDP$ $\square$ Now, let $R$ be the intersection of $KC$ with $AP$. Since $AR$ is tangent to $\Omega$ we get that $90^{\circ} - \angle AKM = 90^{\circ} - \angle APM = \angle PAM = \angle RAC = \angle RKA.$ Therefore $\angle MKR = \angle CKM = 90^{\circ}$, as desired.

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Observe that$$\angle KDP=\angle KCS=\angle KCA=\angle KAP,$$hence $APKD$ is cyclic, so $APKDM$ is cyclic. We now have$$\angle MKA=\angle MPA=90^\circ -\angle PAM=90^\circ -\angle PAC=90^\circ -\angle ABC,$$hence$$\angle CKM=\angle CKA-\angle MKA=(180^\circ -\angle ABC)-(90^\circ -\angle ABC)=90^\circ .$$