Let $$P(4, 4\sqrt{3})$$be a point on the parabola $$y^{2}=4ax$$ and and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to :

Since the parabola is given by $$y^2 = 4ax$$ and the point $$P(4,4\sqrt{3})$$ lies on it, substituting the coordinates into the equation yields: $$(4\sqrt{3})^2 = 4a \cdot 4$$ which simplifies to $$48 = 16a$$ and hence $$a = 3$$. This gives the parabola as $$y^2 = 12x$$.

The focus of this parabola lies at $$(a,0) = (3,0)$$. Since $$PQ$$ is a focal chord, we parametrize the parabola by $$x = a t^2$$ and $$y = 2a t$$; for $$a=3$$ the coordinates become $$(3t^2,6t)$$.

For the point $$P(4,4\sqrt{3})$$ one has $$3t^2 = 4$$ giving $$t^2 = \frac{4}{3}$$ and $$t = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}$$. Substituting the $$y$$-coordinate $$6t = 4\sqrt{3}$$ yields $$t = \frac{2\sqrt{3}}{3}$$, and since the $$y$$-coordinate is positive we take $$t_P = \frac{2\sqrt{3}}{3}$$.

From the condition for focal chords $$t_1 t_2 = -1$$, we obtain for the other end $$t_Q$$ that $$\frac{2\sqrt{3}}{3}\,t_Q = -1$$, so $$t_Q = -\frac{3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$$ after rationalizing. Substituting into the parametric form gives $$x_Q = 3\bigl(-\tfrac{\sqrt{3}}{2}\bigr)^2 = \tfrac{9}{4}$$ and $$y_Q = 6\bigl(-\tfrac{\sqrt{3}}{2}\bigr) = -3\sqrt{3}$$, hence $$Q\bigl(\tfrac{9}{4},-3\sqrt{3}\bigr)$$.

The directrix of the parabola $$y^2 = 4ax$$ is $$x = -a$$, so here it is $$x = -3$$. Since this line is vertical, the feet of the perpendiculars from $$P$$ and $$Q$$ to the directrix lie on horizontal lines through those points. Thus the foot from $$P$$ is $$M(-3,4\sqrt{3})$$ and that from $$Q$$ is $$N(-3,-3\sqrt{3})$$.

Considering the quadrilateral $$PQMN$$ with vertices $$P(4,4\sqrt{3})$$, $$Q\bigl(\tfrac{9}{4},-3\sqrt{3}\bigr)$$, $$N(-3,-3\sqrt{3})$$, and $$M(-3,4\sqrt{3})$$, we observe that $$PQ$$ connects to $$Q$$, then to $$N$$, then to $$M$$, and back to $$P$$. The segments $$QN$$ and $$MP$$ are horizontal and parallel. The length of $$QN$$ is $$\bigl|\tfrac{9}{4} - (-3)\bigr| = \tfrac{21}{4}$$, while the length of $$MP$$ is $$|4 - (-3)| = 7$$. The vertical separation between these parallel sides is $$\bigl|4\sqrt{3} - (-3\sqrt{3})\bigr| = 7\sqrt{3}$$.

Since the area of a trapezoid is half the sum of the parallel sides times the height, we have $$ \text{Area} = \tfrac12\bigl(\tfrac{21}{4} + 7\bigr)\times 7\sqrt{3} = \tfrac12\times \tfrac{49}{4}\times 7\sqrt{3} = \tfrac{343\sqrt{3}}{8}, $$ which matches option D.

As a verification, applying the shoelace formula to the vertices in order $$P(4,4\sqrt{3})$$, $$Q(\tfrac{9}{4},-3\sqrt{3})$$, $$N(-3,-3\sqrt{3})$$, $$M(-3,4\sqrt{3})$$, and back to $$P$$ yields the same result, $$ \text{Area} = \tfrac12\Bigl|\sum(x_i y_{i+1} - y_i x_{i+1})\Bigr| = \tfrac{343\sqrt{3}}{8}. $$ Thus, the area of quadrilateral $$PQMN$$ is $$\tfrac{343\sqrt{3}}{8}$$.

Final Answer: The area of quadrilateral PQMN is $$\tfrac{343\sqrt{3}}{8}$$.