Let perpendicular bisector of $BC$ hit $A'C$ at $E$ then it happens that $\angle MEC=90-\angle BCA'=90-\angle BCA=\angle ABC=\angle BAM$ which means $KAME$ is cyclic, now let $AA' \cap (KAME)=Q' \ne A$ then it happens that $EMAQ'$ is an Isosceles trapezoid therefore $EMA'Q'$ is a parallelogram which means that in favt $Q$ is midpoint of $EQ'$, now as $\angle QQ'K=\angle MEA'=\angle QA'Q'$ which from PoP gives a tangency and thus $QD^2=QQ''^2=QA' \cdot QK$ which gives that $\angle QDK=180-\angle AA'C=180-\angle QA'Q'=180-\angle QQ'K'$ which means that $KDQQ'$ is cyclic and in addition $Q$ is midpoint of arc $DQ'$ here, now as $DQ'$ is parallel to perpendicular bisector of $DM$ we have that it has to be the tangent from $D$ to $(KDQ)$ therefore $PQ$ is tangent to $(KDQ)$ as desired thus we are done .

Trivial AG bash… It comes down to proving that the $y$ coordinate of the circumcenter is the same as $Q$’s.