The diagonal divides the parallelogram into two parts of equal area. In fact, $\triangle SRQ$ and $\triangle QPS$ are congruent (they have matching sides).
Look at $\triangle STQ$ and $\triangle TRQ$. Think of them as having bases $ST$ and $TR$. Then they have the same height. Since $ST=TR$, they also have equal bases. so they have the same area.
Thus our shaded $\triangle STQ$ has area half the area of $SRQ$, which has half the area of the parallelogram. Therefore the area of $\triangle STQ$ is $\frac{1}{2}\cdot \frac{1}{2}$, that is, $\frac{1}{4}$ of the area of the parallelogram.
Remark: The height of the parallelogram cannot be computed from the given information. But we do not need it to find the ratio of the areas.