Solution:

The answer is $\mathbf{(A)}$.

Let $O$ be the center of the triangle. The point $P$ lies by hypothesis on the arc of the circle circumscribed about $AOB$ external to the triangle. The bisector of the angle $A\hat{P}B$ passes through the midpoint of the arc $AB$ opposite to the one on which it lies, which is precisely $O$.

As the lines through $O$ that intersect the segments $AB$ and $AC$ respectively in $X$ and $Y$ vary, the one for which the area of the triangle $AXY$ is minimal is the one parallel to $BC$. Let us call $X'$ and $Y'$ the intersections relative to this latter line and suppose, by symmetry, that $BX < BX'$: we want to show that

$$0 < [AXY] - [AX'Y'] = [OX'X] - [OY'Y]$$

Since these two small triangles have the angle at $O$ equal and $OX' = OY'$, it is enough to observe that $OX > OX' = OY' > OY$.

The minimum possible value for the area of the triangle $AXY$ is therefore

$$\frac{4}{9} \cdot [ABC] = \frac{4}{9} \cdot \frac{\sqrt{3}}{4} = \frac{1}{3 \sqrt{3}}$$

Alternatively. Suppose we fix a Cartesian system with origin at $O$. The points $X$ and $Y$ depend linearly on the slope of the line through $O$, so the area of $AXY$ depends quadratically on this parameter: hence the minimum can only occur at the vertex of the parabola, which by symmetry must correspond to the line parallel to $BC$, or at the endpoints, one of which corresponds to the altitude through $B$. In the first case the area is $4/9$ that of $ABC$, in the other it is half.