The equation of a common tangent to the curves, $$y^2 = 16x$$ and $$xy = -4$$, is:

We begin by assuming that a common tangent to both curves can be written in the slope-intercept form $$y = mx + c$$, where $$m$$ is the slope and $$c$$ is the y-intercept. We will impose the condition of tangency separately on each curve and then combine the resulting relations for $$m$$ and $$c$$.

First, we impose tangency to the parabola $$y^{2}=16x$$. Substituting $$y = mx + c$$ in place of $$y$$ gives

$$\bigl(mx + c\bigr)^{2}=16x.$$

Expanding the left side we get

$$m^{2}x^{2}+2mc\,x+c^{2}-16x=0.$$

This is a quadratic equation in $$x$$. For a line to touch (and not cut) the parabola, this quadratic must have exactly one real root, i.e.\ its discriminant must be zero. The discriminant $$\Delta$$ of the quadratic $$A x^{2}+B x + C = 0$$ is $$\Delta=B^{2}-4AC$$. Here $$A=m^{2},\; B=2mc-16,\; C=c^{2}.$$ Setting $$\Delta=0$$, we obtain

$$(2mc - 16)^{2} - 4 m^{2} c^{2} = 0.$$

Expanding and simplifying term by term,

$$4m^{2}c^{2} - 64mc + 256 - 4m^{2}c^{2} = 0,$$

$$-64mc + 256 = 0,$$

$$mc = 4.$$

Next, we impose tangency to the hyperbola $$xy = -4$$. Substituting $$y = mx + c$$ gives

$$x(mx + c) = -4,$$

$$m x^{2} + c x + 4 = 0.$$

This too is a quadratic in $$x$$. Requiring tangency forces its discriminant to vanish:

$$c^{2} - 4 \cdot m \cdot 4 = 0,$$

$$c^{2} - 16m = 0,$$

$$c^{2} = 16m.$$

We now have a pair of simultaneous relations:

$$mc = 4 \quad \text{and} \quad c^{2} = 16m.$$

From $$mc = 4$$ we can express $$c$$ as $$c = \dfrac{4}{m}.$$ Substituting this into $$c^{2}=16m$$ gives

$$\left(\dfrac{4}{m}\right)^{2}=16m,$$

$$\dfrac{16}{m^{2}}=16m,$$

$$\dfrac{1}{m^{2}} = m,$$

$$m^{3}=1.$$

The real solution of $$m^{3}=1$$ is $$m=1$$ (the other two cubic roots are complex and therefore do not correspond to real tangents). Substituting $$m=1$$ back into $$c = \dfrac{4}{m}$$ we find

$$c = 4.$$

Thus the required common tangent is

$$y = 1\cdot x + 4,$$

or, after rearranging all terms to one side,

$$x - y + 4 = 0.$$

Because this equation matches Option B, we have located the single common tangent correctly.

Hence, the correct answer is Option B.