jmc

To find $\sin \angle PXS$, we build a right triangle with $\angle PXS$ as one of its acute angles. We do so by drawing altitude $\overline{SF}$ from $S$ to diagonal $\overline{PR}$ as shown. We then have $\sin \angle PXS=\sin \angle FXS=\frac{FS}{XS}$.

The Pythagorean Theorem gives us $PR=QS=10$, so $SX=QS/2=5$. We also have $△FPS\sim △SPR$ by AA Similarity (both are right triangles and $\angle SPR=\angle FPS$), so $$\frac{FS}{PS}=\frac{SR}{PR}.$$This gives us $$FS=PS\cdot \frac{SR}{PR}=\frac{6\cdot 8}{10}=\frac{24}{5}.$$Finally, we have $$\sin \angle PXS=\frac{FS}{XS}=\frac{24/5}{5}=\boxed{\frac{24}{25}}.$$