The value of $$\left[\frac{\sqrt{3} +2 \sin P}{1 - 2 \cos P}\right]^3 + \left[\frac{1 + 2 \cos P}{\sqrt{3} - 2 \sin P}\right]^3$$ is:

$$\left[\frac{\sqrt{3} +2 \sin P}{1 - 2 \cos P}\right]^3 + \left[\frac{1 + 2 \cos P}{\sqrt{3} - 2 \sin P}\right]^3=\left[\frac{\frac{\sqrt{3}}{2}+\sin P}{\frac{1}{2}-\cos P}\right]^3+\left[\frac{\frac{1}{2}+\cos P}{\frac{\sqrt{3}}{2}-\sin P}\right]^3$$

$$=\left[\frac{\sin60^{\circ\ }+\sin P}{\cos60^{\circ\ }-\cos P}\right]^3+\left[\frac{\cos60^{\circ\ }+\cos P}{\sin60^{\circ\ }-\sin P}\right]^3$$

$$=\left[\frac{2\sin\left(\frac{60+P}{2}\right)\cos\left(\frac{60-P}{2}\right)}{-2\sin\left(\frac{60+P}{2}\right)\sin\left(\frac{60-P}{2}\right)}\right]^3+\left[\frac{2\cos\left(\frac{60+P}{2}\right)\cos\left(\frac{60-P}{2}\right)}{2\cos\left(\frac{60+P}{2}\right)\sin\left(\frac{60-P}{2}\right)}\right]^3$$

$$=\left[-\cot\left(\frac{60-P}{2}\right)\right]^3+\left[\cot\left(\frac{60-P}{2}\right)\right]^3$$

$$=-\cot^3\left(\frac{60-P}{2}\right)+\cot^3\left(\frac{60-P}{2}\right)$$

$$=0$$

Hence, the correct answer is Option B