Let $\mathrm{AB}$ and $\mathrm{AD}$ be of length $b$ and $h$ respectively. Then the area of $\mathrm{ABCD}=\mathrm{bh}$ and the area of $\mathrm{ABQP}=\frac{1}{2}b(\frac{h}{2}+\frac{h}{3})=\frac{5\mathrm{bh}}{12}$.

Thus the required ratio is $\frac{5}{12}$