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We first note that diagonal $\overline{AC}$ is of length $\sqrt{6}+\sqrt{2}$. It must be that $\overline{AP}$ divides the diagonal into two segments in the ratio $\sqrt{3}$ to $1$. It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions $2\sqrt{3}$ by $\sqrt{3}+1$. The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or $2(\sqrt{3}+1{)}^2-2(\sqrt{3}+1)=2(4+2\sqrt{3})-2\sqrt{3}-2=6+2\sqrt{3}$. The area also includes $4$ circular segments. Two are quarter-circles centered at $P$ of radii $\sqrt{2}$ (the segment bounded by $\overline{PA}$ and $\overline{P{A}^{\prime }}$) and $\sqrt{6}$ (that bounded by $\overline{PC}$ and $\overline{P{C}^{\prime }}$). Assuming $A$ is the bottom-left vertex and $B$ is the bottom-right one, it is clear that the third segment is formed as $B$ swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when $D$ overshoots the final square's left edge. To find these areas, consider the perpendicular from $P$ to $\overline{BC}$. Call the point of intersection $E$. From the previous paragraph, it is clear that $PE=\sqrt{3}$ and $BE=1$. This means $PB=2$, and $B$ swings back inside edge $\overline{BC}$ at a point $1$ unit above $E$ (since it left the edge $1$ unit below). The triangle of the circular sector is therefore an equilateral triangle of side length $2$, and so the angle of the segment is $60^{\circ }$. Imagining the process in reverse, it is clear that the situation is the same with point $D$. The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas $\frac{1}{4}\pi (\sqrt{2}{)}^2-\frac{1}{2}\sqrt{2}\sqrt{2}=\frac{\pi }{2}-1$ and $\frac{1}{4}\pi (\sqrt{6}{)}^2-\frac{1}{2}\sqrt{6}\sqrt{6}=\frac{3\pi }{2}-3$. The other two segments both have area $\frac{1}{6}\pi (2{)}^2-\frac{(2{)}^2\sqrt{3}}{4}=\frac{2\pi }{3}-\sqrt{3}$. The total area is therefore $$(6+2\sqrt{3})+(\frac{\pi }{2}-1)+(\frac{3\pi }{2}-3)+2(\frac{2\pi }{3}-\sqrt{3})$$ $$=2+2\sqrt{3}+2\pi +\frac{4\pi }{3}-2\sqrt{3}$$ $$=\frac{10\pi }{3}+2$$ $$=\frac{1}{3}(10\pi +6)$$ Since $a=10$, $b=6$, and $c=3$, the answer is $a+b+c=10+6+3=\boxed{19}$.