Let $\alpha = AB$ and $\beta = AD$ be the side lengths of the original parallelogram. If you combine the corner triangles obtained from the construction of $PQRS$ then $PBQ + SDR = PBRD$, a parallelogram with side lengths $\frac{\alpha}{3}$ and $\frac{2\alpha}{3}$. Analogously, the other two triangles gives $APS + RCQ = AQCR$ with side lengths $\frac{\alpha}{3}$ and $\beta - \frac{\alpha}{3}$. The area of a parallelogram is given by
$A = \alpha\beta\sin\theta$
where $\theta$ is an angle of the parallelogram. A key observation here is that the angles of the two smaller parallelograms formed are equal to the original. In particular, $PBRD$ contains $\angle ABC$ and $AQCR$ contains $\angle BCD$. Note that this would be enough to reduce the construction to an arbitrary rectangle but not to a square. It turns out that this ratio will depend on the ratio of the side lengths.
Taking the ratio, we have
$\frac{A_{PQRS}}{A_{ABCD}} = \frac{\alpha\beta\sin\theta - \frac{2\alpha^2}{9}\sin\theta - \frac{\alpha}{3}\left(\beta - \frac{\alpha}{3}\right)\sin\theta}{\alpha\beta\sin\theta}$
canceling $\alpha$ and $\sin\theta$ and simplifying gives us
$\frac{A_{PQRS}}{A_{ABCD}} = \frac{\beta - \frac{2\alpha}{9} - \frac{\beta}{3} + \frac{\alpha}{9}}{\beta} = \frac{2}{3} - \frac{1}{9}\cdot\frac{\alpha}{\beta}$
In the case $\alpha = \beta$ this gives us
$\frac{A_{PQRS}}{A_{ABCD}} = \frac{5}{9}$
as required.