Answer: C

Small Hint:

Track the images $A', B', C', D';$ the swept region is four circular sectors plus four triangles

Big Hint:

$AP = \sqrt{2}$ and $PC = \sqrt{6}$ give the sector radii, and $\triangle BPH$ is a $30$–$60$–$90$ triangle

Solution:

Let $A', B', C', D'$ be the images of the vertices under the rotation. The swept region decomposes into four circular sectors and four triangles.

Since $AP = \sqrt{2}$ and $PC = AC - AP = \sqrt{6},$ the sectors at $A$ and $C$ have areas $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}.$ If $H$ is the midpoint of $AA',$ then $PH=AH=1$ and $HB=\sqrt3,$ so $\triangle BPH$ is a $30$-$60$-$90^\circ$ triangle and $PB=2.$ Hence the two $60^\circ$ sectors along $BC$ each have area $\frac{2\pi}{3}.$ The two triangles with altitude $PH$ contribute $\sqrt3-1,$ and the other congruent pair has altitude $\sqrt3$ and contributes $3-\sqrt3.$ Thus the four triangles contribute $2.$

The total area is $$\begin{gathered} \dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 \\ = \dfrac{10\pi + 6}{3}, \end{gathered}$$ so $a + b + c = 10 + 6 + 3 = 19.$

Thus, the correct answer is C.