China Mathematical Competition

$O$ and $H$, respectively. Then $O$ is the centre of the base square, $H$ lies on $OM$, and $\frac{GH}{PO} = \frac{HM}{OM} = \frac{GM}{PM} = \frac{1}{3}$.

For the sake of convenience, let $AB = PO = 6$. Thus, $GH = \frac{PO}{3} = 2$, $OH = \frac{2}{3}OM = 2$. And since $AO = 3\sqrt{2}$, $\angle AOH = 135^\circ$, we have
$$AH^2 = AO^2 + OH^2 - 2AO \cdot OH \cdot \cos \angle AOH = 34,$$
and thus $AG = \sqrt{AH^2 + GH^2} = \sqrt{38}$.

The angle between line $AG$ and base $ABCD$ is equal to $\angle GAH$. Therefore, the desired sine is
$$\sin \angle GAH = \frac{GH}{AG} = \frac{2}{\sqrt{38}} = \frac{\sqrt{38}}{19}.$$