jmc

Denote the midpoint of $\overline{DC}$ be $E$ and the midpoint of $\overline{AB}$ be $F$. Because they are the circumcenters, both Os lie on the perpendicular bisectors of $AB$ and $CD$ and these bisectors go through $E$ and $F$. It is given that $\angle O_{1}P{O}_2=120^{\circ }$. Because $O_{1}P$ and $O_{1}B$ are radii of the same circle, the have the same length. This is also true of $O_{2}P$ and $O_{2}D$. Because $m\angle CAB=m\angle ACD=45^{\circ }$, $m\stackrel{⌢}{\mathrm{PD}}=m\stackrel{⌢}{\mathrm{PB}}=2(45^{\circ })=90^{\circ }$. Thus, $O_{1}PB$ and $O_{2}PD$ are isosceles right triangles. Using the given information above and symmetry, $m\angle DPB=120^{\circ }$. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees. Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles $O_{1}BF$ and $O_{2}DE$ have measures of 30 degrees. Thus, both triangles $O_{1}BF$ and $O_{2}DE$ are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, $D{O}_2=B{O}_1=4\sqrt{3}$. Because of 45-45-90 right triangles, $PB=PD=4\sqrt{6}$. Now, letting $x=AP$ and using Law of Cosines on $△ABP$, we have $$96=144+x^2-24x\frac{\sqrt{2}}{2}$$$$0=x^2-12x\sqrt{2}+48$$ Using the quadratic formula, we arrive at $$x=\sqrt{72}\pm \sqrt{24}$$ Taking the positive root, $AP=\sqrt{72}+\sqrt{24}$ and the answer is thus $\boxed{96}$.