9. As shown in Figure 1, in the tetrahedron $P-A B C$, the lateral face $P A C \perp$ the base $A B C$, the base $A B C$ is an equilateral triangle with a side length of 1, $P A=P C$, $\angle A P C=90^{\circ}$, and $M$ is the midpoint of edge $B C$. Then the distance between $A B$ and $P M$ is ( ).
(A) $\frac{\sqrt{3}}{4}$
(B) $\frac{1}{2}$
(C) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{3}}{3}$
9. A.

As shown in Figure 3, draw $PO \perp AC$ at point $O$, then $O$ is the midpoint of $AC$. Connect $OM$.

Since $M$ is the midpoint of $BC$, we know that $OM \parallel AB$.

Draw $MN \perp AB$ at point $N$.
It is easy to see that $MN = \frac{\sqrt{3}}{4}$.

Since $OM \parallel AB$, therefore, $MN \perp OM$.
Also, since the lateral face $PAC \perp$ the base face $ABC$, and $PO \perp AC$, then $PO \perp$ the base face $ABC$, and $PO \perp MN$.
Thus, $MN \perp$ the plane $POM$, and $MN \perp PM$.
Therefore, $MN$ is the common perpendicular of $AB$ and $PM$.
[Note] This problem can also be solved using vector methods.