A Pyramid over a Parallelogram Base

As shown in the figure, in the pyramid $P\text{-}ABCD$ the base $ABCD$ is a parallelogram with $\angle ABC=120^\circ$, $AB=1$, $BC=4$, and $PA=\sqrt{15}$. The points $M$ and $N$ are the midpoints of $BC$ and $PC$ respectively, and $PD\perp DC$, $PM\perp MD$.

1. Prove that $AB\perp PM$.

2. Find the sine of the angle between the line $AN$ and the plane $PDM$.