9. As shown in Figure 3, in rhombus $A B C D$, $\angle A=100^{\circ}$, $M$ and $N$ are the midpoints of sides $A B$ and $B C$ respectively, and $M P \perp C D$ at point $P$. Then the degree measure of $\angle N P C$ is $\qquad$
$=9.50^{\circ}$.
Since quadrilateral $ABCD$ is a rhombus, therefore,
$$
AB=BC, \angle B=180^{\circ}-\angle A=80^{\circ} \text {. }
$$

Also, $M$ and $N$ are the midpoints of sides $AB$ and $BC$, respectively, so $BM=BN$,
$$
\angle BMN=\angle BNM=\frac{1}{2}\left(180^{\circ}-80^{\circ}\right)=50^{\circ} \text {. }
$$

Let the midpoint of $MP$ be
$E$, as shown in Figure 10, connect $NE$. Then
$NE \parallel BM \parallel CP$.
Since $MP \perp CD$, then $NE \perp MP$, so $NM=NP$.

Therefore, $\angle NMP$ $=\angle NPM$.
Thus $\angle NPC=\angle NMB=50^{\circ}$.