Solution:

Place $A = (0,0)$ and $B = (1,0)$ on the unit square. $AP$ is the middle length when $BP \lt AP \lt AB$ or $AB \lt AP \lt BP.$ These two conditions carve out regions bounded by the circle of radius $1$ centered at $A$ (where $AP = AB$) and the perpendicular bisector $x = \tfrac12$ (where $AP = BP$). Let $S$ be where that circle meets $x = \tfrac12.$ Then $\triangle ABS$ is equilateral, so $\angle QAS = 60^\circ.$ Working out the pieces, the larger region has area $\frac{4\pi - 3\sqrt3}{24}$ and the smaller $\frac{12 - 2\pi - 3\sqrt3}{24}.$ They add to $\frac{6 + \pi - 3\sqrt3}{12}.$ So $a + b + c + d + e = 6 + 1 + 3 + 3 + 12 = 25.$ Thus, A is the correct answer.