To solve this problem, let's consider the geometry and algebra involved:

1. Concept of Area Section: When a line parallel to the base of a triangle divides it into two parts of equal areas, it divides the height in a specific ratio. For a triangle with height $h$, this line is at height $\frac{h}{\sqrt{2}}$.

2. Using Similar Triangles: Since $MN \parallel PR$, triangle $\triangle PMN$ is similar to triangle $\triangle PQR$. The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

   Area ratio: $\left(\frac{PM}{PQ}\right)^2 = \frac{1}{2}$

3. Solve for $\frac{PM}{PQ}$:

   $\left(\frac{PM}{PQ}\right)^2 = \frac{1}{2}$

   $\frac{PM}{PQ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

So the correct option is $\frac{\sqrt{2}}{2}$, which is not explicitly listed, indicating the answer should be (E) None of these.