Let $$S = \{\theta \in [0, 2\pi] : 8^{2\sin^2\theta} + 8^{2\cos^2\theta} = 16\}$$. Then $$n(S) + \displaystyle\sum_{\theta \in S} \left(\sec\left(\dfrac{\pi}{4} + 2\theta\right) \csc\left(\dfrac{\pi}{4} + 2\theta\right)\right)$$ is equal to:

We need to find $$n(S) + \displaystyle\sum_{\theta \in S} \sec\left(\frac{\pi}{4} + 2\theta\right)\csc\left(\frac{\pi}{4} + 2\theta\right)$$, where $$S = \{\theta \in [0, 2\pi] : 8^{2\sin^2\theta} + 8^{2\cos^2\theta} = 16\}$$.

Let $$t = 2\sin^2\theta$$, so $$2\cos^2\theta = 2 - t$$.

$$8^t + 8^{2-t} = 16$$

Let $$u = 8^t$$: $$u + \frac{64}{u} = 16$$, so $$u^2 - 16u + 64 = 0$$, giving $$(u - 8)^2 = 0$$.

Therefore $$u = 8$$, so $$8^t = 8^1$$, giving $$t = 1$$, i.e., $$\sin^2\theta = \frac{1}{2}$$.

$$\sin\theta = \pm\frac{1}{\sqrt{2}}$$, so $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$.

$$n(S) = 4$$

$$\sec\alpha \cdot \csc\alpha = \frac{1}{\sin\alpha \cos\alpha} = \frac{2}{\sin 2\alpha}$$, where $$\alpha = \frac{\pi}{4} + 2\theta$$.

For each $$\theta$$, we compute $$\sin(2\alpha) = \sin\left(\frac{\pi}{2} + 4\theta\right) = \cos(4\theta)$$:

$$\theta = \frac{\pi}{4}: \cos(\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{3\pi}{4}: \cos(3\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{5\pi}{4}: \cos(5\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{7\pi}{4}: \cos(7\pi) = -1 \implies \text{value} = -2$$

Sum = $$-2 - 2 - 2 - 2 = -8$$

$$n(S) + \text{Sum} = 4 + (-8) = -4$$

Therefore, the correct answer is Option C: $$-4$$.