Let $AH$ meet $PQ$ at $K$
$\angle MAP = \angle MAB + 60 = B+60 \implies \angle NAP = 180-B-60=120-B$
Similarly, $\angle NAQ=120-C$
Hence, $\frac{PN}{NQ} = \frac{AP}{AQ}.\frac{sin(120-B)}{sin(120-C)}$
$\angle HAP = \angle HAB + 60 = C+60 \implies \angle KAP = 180-C-60=120-C$
Similarly, $\angle KAQ=120-B$
Hence, $\frac{PK}{KQ} = \frac{AP}{AQ}.\frac{sin(120-C)}{sin(120-B)}$
Therefore, $\frac{PN}{NQ}.\frac{PK}{KQ} = \frac{AP^2}{AQ^2}$
$\angle HAP=\angle HCQ = C+60$
$\frac {HC}{CQ}=\frac{HC}{AC}=cos C$
$\frac {AH}{AP}=\frac{AH}{AB}=cos C \implies \frac {HC}{CQ}=\frac {AH}{AP}$
Hence, $\triangle HAP \sim \triangle HCQ \implies \frac{AP}{CQ} = \frac{PH}{HQ} \implies \frac{AP}{AQ}=\frac{PH}{HQ}$
Therefore, $\frac{AP^2}{AQ^2}=\frac{PH^2}{HQ^2} \implies \frac{PN}{NQ}.\frac{PK}{KQ}=\frac{PH^2}{HQ^2}$
Therefore, $HK, HN$ are isogonal conjugates through $H$ in $\triangle PHQ$
Hence, $\angle NHP = \angle KHQ \implies \angle NHP = \angle AHQ$