I guess something's wrong with the diagram, can someone confirm??

You are correct, there is a typo in the second fraction. Instead of $\frac{DP}{DN}=\frac{BC}{CD}$ , the correct is $\frac{DP}{DN}=\frac{BD}{CD}$ . My original post with the typo was parmenides51 wrote: Let the inscribed circle of the triangle $\vartriangle ABC$ touch side $BC$ at $M$ , side $CA$ at $N$ and side $AB$ at $P$ . Let $D$ be a point from $\left[ NP \right]$ such that $\frac{DP}{DN}=\frac{BC}{CD}$ . Show that $DM \perp PN$ . Thanks

Edit: Just realised that there was a much better solution, and @below may this solution help you, Now, $\angle BPN=\angle PNC \implies \Delta DNC \sim \Delta DPB$, Hence, $\frac{BD}{DC}=\frac{PB}{CN}=\frac{BM}{CM} \implies \angle BDM=\angle CDM \implies \boxed{ MD \perp PN}$

Can you explain $BP||CN$?

in alastormoody's solution, triangles dnc and dpb are not similar, because that would mean SSA rt?