jmc

Since $PQ$ is parallel to $SR,$ the height of $△PQS$ (considering $PQ$ as the base) and the height of $△SRQ$ (considering $SR$ as the base) are the same (that is, the vertical distance between $PQ$ and $SR$).

Since $SR$ is twice the length of $PQ$ and the heights are the same, the area of $△SRQ$ is twice the area of $△PQS.$ In other words, the area of $△PQS$ is $\frac{1}{3}$ of the total area of the trapezoid, or $\frac{1}{3}\times 12=\boxed{4}.$