Japan Mathematical Olympiad

$$\frac{6+7\sqrt{3}}{2}$$
Take a point $Q$ in the opposite side from the point $C$ with respect to the line $AB$ so as to make the triangles $ABQ$ and $ACP$ similar. The similarity ratio of the triangle $ABQ$ to the triangle $ACP$ is given by $AB : AC = 2 : 1$, and so we obtain $AQ = 2AP = 2\sqrt{3}$, $BQ = 2CP = 4$. Also, from $\angle QAB = \angle PAC$ we get
$$\angle QAP = \angle QAB + \angle BAP = \angle PAC + \angle BAP = \angle BAC = 60^{\circ},$$
which, together with the fact that $AQ : AP = 2 : 1$, yield $\angle APQ = 90^{\circ}$ and $PQ = \sqrt{3}AP = 3$.

From these facts, we deduce that $BP^2 = BQ^2 + PQ^2$, and this in turn implies that $\angle BQP = 90^{\circ}$. Consequently, we get $AB^2 = PQ^2 + (AP + BQ)^2 = 28 + 8\sqrt{3}$, and the area of the triangle $ABC$ is given by
$$\frac{1}{2} \cdot AB \cdot AC \cdot \sin 60^{\circ} = \frac{\sqrt{3}}{8} AB^2 = \frac{6+7\sqrt{3}}{2}$$