If $$f(x) = \begin{cases} x + a, & x \le 0 \\ |x - 4|, & x > 0 \end{cases}$$ and $$g(x) = \begin{cases} x + 1, & x < 0 \\ (x-4)^2 + b, & x \ge 0 \end{cases}$$ are continuous on $$\mathbb{R}$$, then $$(gof)(2) + (fog)(-2)$$ is equal to:

We have $$f(x) = \begin{cases} x + a, & x \le 0 \\ |x - 4|, & x > 0 \end{cases}$$ and $$g(x) = \begin{cases} x + 1, & x < 0 \\ (x-4)^2 + b, & x \ge 0 \end{cases}$$

To ensure continuity at $$x = 0$$, for $$f$$ the left limit is $$0 + a = a$$ and the right limit is $$|0 - 4| = 4$$, so $$a = 4$$. Substituting this into $$g$$, the left limit is $$0 + 1 = 1$$ and the right value is $$(0-4)^2 + b = 16 + b$$, giving $$b = -15$$.

Next, since $$2 > 0$$, we have $$f(2) = |2 - 4| = 2$$ and then $$g(f(2)) = g(2) = (2 - 4)^2 + (-15) = 4 - 15 = -11$$ (since $$2 \ge 0$$).

Then, for $$(f \circ g)(-2)$$, we compute $$g(-2) = -2 + 1 = -1$$ (since $$-2 < 0$$) and $$f(g(-2)) = f(-1) = -1 + 4 = 3$$ (since $$-1 \le 0$$).

From this, $$(g \circ f)(2) + (f \circ g)(-2) = -11 + 3 = -8$$.

Therefore, the correct answer is Option D: $$-8$$.