To determine the principal value of $$\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$$, we need to understand the properties of the inverse sine function, also known as the arcsine function.

The sine function, $$\sin(x)$$, is periodic with period $$2\pi$$, meaning that for any integer $$k$$, we have:

$$\sin(x) = \sin(x + 2k\pi)$$.

The inverse sine function, $$\sin^{-1}(x)$$, has a restricted domain, typically taken as:

$$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$,

where $$y = \sin^{-1}(x)$$.

Given the expression inside the inverse sine function:

$$\sin\left(\frac{2\pi}{3}\right)$$,

we know:

$$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$.

Now, we want to find the principal value of $$\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$$.

Let:

$$x = \frac{2\pi}{3}$$,

so:

$$\sin\left(\frac{2\pi}{3}\right) = \sin(x) = \frac{\sqrt{3}}{2}$$.

To find the principal value, we need to find an angle $$y$$ within the range $$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$ such that $$\sin(y) = \frac{\sqrt{3}}{2}$$.

The principal value that satisfies this condition is:

$$y = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$$.

Given the range of the inverse sine function, the corresponding angle is:

$$y = \frac{\pi}{3}$$.

Now, we need to match this value with one of the given options. Since there is no option for $$\frac{\pi}{3}$$ directly, but recalling the properties of the sine function and the range, and knowing that $$\frac{2\pi}{3}$$ is not within the range of the principal values, we need to reconsider equivalent values.

None of the given options match exactly the principal value we calculated, which is $$\frac{\pi}{3}$$. Therefore, the correct answer is:

Option D: none