$\triangle MDB$ similar $\triangle PDN$ $\rightarrow$ $\triangle KNM$ is isosceles $\rightarrow$$\triangle PMK$ is isosceles, $PM=PK$

We have $\angle BPN=\angle BKN=90^\circ-\angle KBL=90^\circ-\angle PBK=\angle KMN$, so $PN\perp BK$ and $NM=NK$. Therefore, $PN$ is the perpendicular bisector of $KM$, and $PM=PK$, so $KPM$ is isosceles. A corollary is that $M$ is the orthocenter of $PBN$.

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