Denote $R=AP\cap MN$. See, that $\measuredangle MAB = \measuredangle DAN$. Hence from DDIT applied to $NPMC$ and point $A$. As $(AM,AN)$, $(AD,AB)$ are isogonal pairs in $NAB$ we get that $(AP, AC)$ is an isogonal pair. Hence $\measuredangle PAM=\measuredangle NAC$. From right angles is $AMCP$ concyclic, hence $\measuredangle NMC = \measuredangle NAC$. From this we get that triangles $ARM$ and $ANC$ are similar, hence $\measuredangle ARM=90^\circ$. $\square$

Note that by the isogonality lemma, since $AB, AD$ are isogonal in $\angle MAN$, $AC, AP$ are isogonal in $\angle MAN$. Since $C$ is the antipode of $A$ in $(MAN)$, $AP$ is perpendicular to $MN$, as required.