Let the focal chord of the parabola $$P: y^2 = 4x$$ along the line $$L: y = mx + c, m > 0$$ meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola $$H: x^2 - y^2 = 4$$. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is

We have the parabola $$P: y^2 = 4x$$ with vertex $$O(0,0)$$ and focus at $$(1,0)$$. A focal chord lies along the line $$L: y = mx + c$$ with $$m > 0$$. Since $$L$$ passes through the focus $$(1,0)$$, we get $$0 = m + c$$, so $$c = -m$$. Thus $$L: y = m(x - 1)$$.

We are also told that $$L$$ is tangent to the hyperbola $$H: x^2 - y^2 = 4$$, where $$a^2 = 4$$ and $$b^2 = 4$$. For a line $$y = mx + c$$ to be tangent to the hyperbola $$x^2 - y^2 = 4$$, the condition is $$c^2 = a^2 m^2 - b^2 = 4m^2 - 4$$. Since $$c = -m$$, we have $$m^2 = 4m^2 - 4$$, which gives $$3m^2 = 4$$, so $$m^2 = \dfrac{4}{3}$$ and $$m = \dfrac{2}{\sqrt{3}}$$.

Now we find the points $$M$$ and $$N$$ where $$L$$ meets the parabola. Substituting $$y = m(x-1)$$ into $$y^2 = 4x$$: $$m^2(x-1)^2 = 4x$$. With $$m^2 = 4/3$$: $$\dfrac{4}{3}(x-1)^2 = 4x$$, so $$(x-1)^2 = 3x$$, giving $$x^2 - 5x + 1 = 0$$.

The roots are $$x_1 = \dfrac{5 + \sqrt{21}}{2}$$ and $$x_2 = \dfrac{5 - \sqrt{21}}{2}$$. The corresponding $$y$$-values are $$y_i = m(x_i - 1)$$. We note that $$x_1 > 1$$ so $$y_1 > 0$$, and $$x_2 < 1$$ so $$y_2 < 0$$. Hence $$M$$ and $$N$$ lie on opposite sides of the $$x$$-axis.

Now, the focus of the hyperbola on the positive $$x$$-axis is $$F$$. Since $$a^2 = b^2 = 4$$, we have $$c_H = \sqrt{a^2 + b^2} = \sqrt{8} = 2\sqrt{2}$$, so $$F = (2\sqrt{2}, 0)$$.

The quadrilateral $$OMFN$$ has vertices $$O(0,0)$$, $$M$$, $$F(2\sqrt{2}, 0)$$, and $$N$$. Since $$O$$ and $$F$$ both lie on the $$x$$-axis, and $$M$$, $$N$$ are on opposite sides of it, the area of the quadrilateral equals $$\dfrac{1}{2} \cdot OF \cdot (|y_M| + |y_N|)$$.

We compute $$|y_M| + |y_N| = m|x_1 - 1| + m|1 - x_2| = m\left(\dfrac{3 + \sqrt{21}}{2} + \dfrac{\sqrt{21} - 3}{2}\right) = m \cdot \sqrt{21}$$.

With $$m = \dfrac{2}{\sqrt{3}}$$: $$|y_M| + |y_N| = \dfrac{2\sqrt{21}}{\sqrt{3}} = 2\sqrt{7}$$.

Hence the area is $$\dfrac{1}{2} \cdot 2\sqrt{2} \cdot 2\sqrt{7} = 2\sqrt{14}$$.

Hence, the correct answer is Option B.