Clearly $\widehat{ANB}=90^{\circ}+\tfrac{1}{2}\widehat{ACB}$ means that $N$ is on circle $\omega$ through $A,B$ and tangent to $CA,CB.$ Hence $\widehat{PAN}=\tfrac{1}{2}\widehat{CAN}=\tfrac{1}{2}\widehat{NBA}=\widehat{NBP}$ $\Longrightarrow$ $P$ is midpoint of the arc $NA$ of $\omega$ $\Longrightarrow$ $P$ is on perpendicular bisector of $\overline{AN}.$ $CD \parallel AN$ gives $\widehat{CDB}=\widehat{AND}=90^{\circ}-\tfrac{1}{2}\widehat{ACB}=\widehat{CAB}$ $\Longrightarrow$ $D \in \odot(ABC)$ $\Longrightarrow$ $\widehat{ADN}=\widehat{ADB}=\widehat{ACB}=90^{\circ}-\tfrac{1}{2}\widehat{DNA}$ $\Longrightarrow$ $\triangle DAN$ is D-isosceles, therefore $DP$ is perpendicular bisector of $\overline{AN}$ $\Longrightarrow$ $DP \perp AN,$ as desired.