Let $$S = \{x \in [-\pi, \pi] : \sin x(\sin x + \cos x) = a, a \in \mathbb{Z}\}$$. Then $$n(S)$$ is equal to :

We expand the given expression:

$$\sin x(\sin x + \cos x) = \sin^2 x + \sin x \cos x = \frac{1 - \cos 2x}{2} + \frac{\sin 2x}{2}$$

So the equation becomes $$\sin 2x - \cos 2x = 2a - 1$$, which can be written as:

$$\sqrt{2}\sin\left(2x - \frac{\pi}{4}\right) = 2a - 1$$

For real solutions, we need $$|2a - 1| \leq \sqrt{2}$$. Since $$a \in \mathbb{Z}$$, this gives $$\frac{1 - \sqrt{2}}{2} \leq a \leq \frac{1 + \sqrt{2}}{2}$$, i.e., $$-0.207 \leq a \leq 1.207$$. So $$a = 0$$ or $$a = 1$$.

Case $$a = 0$$: We solve $$\sin\left(2x - \frac{\pi}{4}\right) = \frac{-1}{\sqrt{2}}$$.

This gives $$2x - \frac{\pi}{4} = -\frac{\pi}{4} + 2k\pi$$ or $$2x - \frac{\pi}{4} = \pi + \frac{\pi}{4} + 2k\pi$$.

From the first: $$x = k\pi$$. In $$[-\pi, \pi]$$: $$x = -\pi, 0, \pi$$.

From the second: $$x = \frac{3\pi}{4} + k\pi$$. In $$[-\pi, \pi]$$: $$x = -\frac{\pi}{4}, \frac{3\pi}{4}$$.

So 5 solutions for $$a = 0$$.

Case $$a = 1$$: We solve $$\sin\left(2x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$.

This gives $$2x - \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi$$ or $$2x - \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi$$.

From the first: $$x = \frac{\pi}{4} + k\pi$$. In $$[-\pi, \pi]$$: $$x = -\frac{3\pi}{4}, \frac{\pi}{4}$$.

From the second: $$x = \frac{\pi}{2} + k\pi$$. In $$[-\pi, \pi]$$: $$x = -\frac{\pi}{2}, \frac{\pi}{2}$$.

So 4 solutions for $$a = 1$$.

The total number of elements in $$S$$ is $$5 + 4 = 9$$.

Hence, the correct answer is Option 4.