Solution:

Group by Euler's formula: $P(x) = 1 + e^{ix} - e^{2ix} + e^{3ix}.$ The imaginary part is $\sin x - \sin 2x + \sin 3x = (\sin x + \sin 3x) - \sin 2x = \sin 2x(2\cos x - 1).$

This vanishes when $\sin 2x = 0$ (so $x = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}$) or $\cos x = \tfrac12$ (so $x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$).

Checking the real part $1 + \cos x - \cos 2x + \cos 3x$ at each of these values gives $\pm 2$ or $1,$ never $0.$ So no $x$ makes $P(x) = 0.$

Thus, the correct answer is A.