Note: The figure shown is representative.

Let side length of the square $PQRS$ be $a = 2\text{ cm}$.

1. Let $\angle LPQ = \theta$.

2. In right-angled triangle $

riangle PQL$(since$\angle PQL = 90^\circ$):

3. Since $\angle QPS = 90^\circ$ and $\angle LPN = 90^\circ$:

   $$\angle LPS = 90^\circ - \theta$$

   $$\angle SPN = \angle LPN - \angle LPS = 90^\circ - (90^\circ - \theta) = \theta$$

4. Since $S$ lies on the line segment $MN$ and $PN \perp MN$, $

riangle PNS$is a right-angled triangle at$N$with hypotenuse$PS = 2$\text{ cm}$$:

5. The area of the rectangle $PLMN$ is given by:

   $$\text{Area} = PL \times PN = \left(\frac{2}{\cos\theta}\right) \times (2\cos\theta) = 4\text{ cm}^2$$

Thus, the area is independent of $\theta$ and equals 4, which corresponds to option (D).