Solution:

Take vectors centered at the center $O$ of the triangle. Write the vector $OA$ as $\mathbf{A}$ etc. Then
$$PA^{2} + PB^{2} + PC^{2} = (\mathbf{P} - \mathbf{A})^{2} + (\mathbf{P} - \mathbf{B})^{2} + (\mathbf{P} - \mathbf{C})^{2} = 3 P^{2} + (A^{2} + B^{2} + C^{2}) - 2 \mathbf{P} \cdot (\mathbf{A} + \mathbf{B} + \mathbf{C}) = 15 P^{2},$$
since $A^{2} = B^{2} = C^{2} = 4 P^{2}$ and $\mathbf{A} + \mathbf{B} + \mathbf{C} = 0$. Finally, the side is $2$, so an altitude is $\sqrt{3}$ and the inradius is $\sqrt{3}/3 = 1/\sqrt{3}$, so $PA^{2} + PB^{2} + PC^{2} = 15/3 = 5$.

Take $Q$ outside the triangle so that $BQ = BP$ and $CQ = AP$. Then $BQC$ and $BPA$ are congruent, so $\angle ABP = \angle CBQ$ and hence $\angle PBQ = 60^{\circ}$, so $PBQ$ is equilateral. Hence $PQ$ is $PB$ and $PQC$ has sides equal to $PA$, $PB$, $PC$. If we construct two similar points outside the other two sides then we get a figure with total area equal to $2$ area $ABC$ and to $3$ area $PQC$ plus area of three equilateral triangles sides $PA$, $PB$, $PC$. Hence $3$ area $PQC = 2$ area $ABC$ $-$ area $ABC (PA^{2} + PB^{2} + PC^{2}) / PA^{2} = (3/4)$ area $ABC = (3\sqrt{3})/4$. So area $PQC = (\sqrt{3})/4$.