It appears to be $\frac 1{16}$. We have an equality for a square and the inequality holds for rectangles. When the polygon is concave, we can easily extend its border so that it encompasses more squares but the perimeter stays the same or decreases, so for a certain positive integer $n$, the polygon with perimeter $\le n$ and maximal area must be convex, and since the polygons we're working with have right angles, it must be a rectangle. This means that the ratio $\frac{S(F}{P^2(F)}$ is smaller for a concave figure than for a certain rectangle, and that of the rectangle is $\le\frac 1{16}$, which means that $\frac 1{16}$ is Ok.

Only simpler solution.Consider it's circumscribed rectangle .It's Area is bigger and it's perimeter is smaller.Inequality is obvious for rectangles.

That's exactly what I thought, but I thought we needed some justification for the fact that the perimeter of the circumscribed rectangle is smaller (you can't just say it's smaller ).