2. In the quadrilateral pyramid $P-ABCD$, it is known that quadrilateral $ABCD$ is a rectangle, and $AB=4, BC=3, PA=PB=PC=PD=5$, $AC$ intersects with $BD$ at point $O$, and $M$ is the midpoint of side $PC$. Then the angle between $OM$ and the plane $PBC$ is
2. $\arcsin \frac{4 \sqrt{273}}{91}$.

Take the midpoint $E$ of side $BC$, then $OE \perp BC$, $PE \perp BC$.
Thus, $BC \perp$ plane $POE$.
Draw $OF \perp PE$ through point $O$, then $OF \perp$ plane $PBC$. Therefore, $\angle OMF$ is the angle we need.
Since $OM$ is the median of the hypotenuse of right $\triangle POC$, we have $OM=\frac{5}{2}$.
Also, $OF=\frac{OP \cdot OE}{PE}=\frac{10 \sqrt{3}}{\sqrt{91}}$, thus,
$$
\sin \angle OMF=\frac{OF}{OM}=\frac{4 \sqrt{3}}{\sqrt{91}} \text{. }
$$

Hence, the angle between $OM$ and plane $PBC$ is $\arcsin \frac{4 \sqrt{273}}{91}$.