It's Geometry !!!

I got the idea of making this problem from This One.

$\triangle PCD$ is equilateral. See the proof here : Equilateral Triangle inside a Square

So, $PC=PD=CD=4$

In $\triangle CNF$ , $\cos FCN = \frac{CF}{CN} \Rightarrow \cos 30 = \frac{2}{CN} \Rightarrow \frac{\sqrt{3}}{2} = \frac{2}{CN} \Rightarrow CN = \frac{4}{\sqrt{3}}$

$PN = CP - CN = 4 - \frac{4}{\sqrt{3}}$

As $MN$ is parallel to $CD$ , $\triangle PMN$ is also equilateral.

$MN = PM = PN = 4 - \frac{4}{\sqrt{3}}$

$\angle MQN = 2 \angle MPN = 120^\circ$

$$

$\angle MQN + \angle MPN = 180^\circ$ . Quadrilateral $PMQN$ is cyclic.

Now using Ptolemy's Theorem ,

$PM \times QN + PN \times QM = MN \times PQ$

$\Rightarrow PM \times QN + PM \times QM = PM \times PQ$ $[$ As $PM = PN = MN]$

$\Rightarrow QN + QM = PQ$

$\Rightarrow \frac{3\sqrt{5}+8\sqrt{3} - 8}{2\sqrt{3}} - MN = PQ$ $[$ Given $MN + MQ + NQ = \frac{3\sqrt{5}+8\sqrt{3} - 8}{2\sqrt{3}}]$

$\Rightarrow \frac{3\sqrt{5}+8\sqrt{3} - 8}{2\sqrt{3}} - (4 - \frac{4}{\sqrt{3}}) = PQ$

$\Rightarrow PQ = \frac{\sqrt{15}}{2}$

$\Rightarrow PQ^2 = \frac{15}{4}$

$$

$15+4=\boxed{19}$