Let $$p, q \in \mathbb{R}$$ and $$(1 - \sqrt{3}i)^{200} = 2^{199}(p + iq)$$, $$i = \sqrt{-1}$$. Then $$p + q + q^2$$ and $$p - q + q^2$$ are roots of the equation.

We need to find the equation whose roots are $$p + q + q^2$$ and $$p - q + q^2$$.

$$ 1 - \sqrt{3}i = 2\left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 2(\cos(-60°) + i\sin(-60°)) = 2e^{-i\pi/3} $$ $$ (1 - \sqrt{3}i)^{200} = 2^{200} e^{-i \cdot 200\pi/3} $$

$$\frac{200}{3} = 66\frac{2}{3}$$, so $$200\pi/3 = 66\pi + 2\pi/3$$.

$$ e^{-i \cdot 200\pi/3} = e^{-i(66\pi + 2\pi/3)} = e^{-i \cdot 2\pi/3} $$

(since $$e^{-i \cdot 66\pi} = e^{0} = 1$$ as 66 is even)

$$ = \cos(-2\pi/3) + i\sin(-2\pi/3) = -\frac{1}{2} - \frac{\sqrt{3}}{2}i $$ $$ (1-\sqrt{3}i)^{200} = 2^{200}\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 2^{199}(-1 - \sqrt{3}i) $$

Given $$(1-\sqrt{3}i)^{200} = 2^{199}(p + iq):$$

$$ p = -1, \quad q = -\sqrt{3} $$ $$ p + q + q^2 = -1 + (-\sqrt{3}) + 3 = 2 - \sqrt{3} $$ $$ p - q + q^2 = -1 - (-\sqrt{3}) + 3 = 2 + \sqrt{3} $$

Sum of roots:

$$ (2-\sqrt{3}) + (2+\sqrt{3}) = 4 $$

Product of roots:

$$ (2-\sqrt{3})(2+\sqrt{3}) = 4 - 3 = 1 $$

Equation:

$$ x^2 - 4x + 1 = 0 $$

The correct answer is Option B:

$$ x^2 - 4x + 1 = 0 $$