The equation \( z^{99} = 1 \) describes the 99th roots of unity. The roots of this equation can be expressed in the form:
\[ z_k = e^{\frac{2\pi i k}{99}} \quad \text{for } k = 0, 1, 2, \ldots, 98. \]
Each root \( z_k \) has real and imaginary parts given by:
\[ z_k = x_k + iy_k = \cos\left(\frac{2\pi k}{99}\right) + i \sin\left(\frac{2\pi k}{99}\right). \]
Thus, we have:
\[ x_k = \cos\left(\frac{2\pi k}{99}\right), \quad y_k = \sin\left(\frac{2\pi k}{99}\right). \]
The task requires us to construct the polynomial \( P(z) \), which has roots \( x_k - iy_k \). Writing the roots of \( P(z) \) explicitly, we have:
\[ x_k - iy_k = \cos\left(\frac{2\pi k}{99}\right) - i \sin\left(\frac{2\pi k}{99}\right) = e^{-\frac{2\pi i k}{99}}. \]
In other words, the roots of \( P(z) \) are exactly the 99th roots of unity, specifically the same set of points just traversed in reverse order around the unit circle.
Given that the roots of \( P(z) \) are the 99th roots of unity, we observe that the polynomial \( P(z) \) can be expressed as:
\[ P(z) = (z - e^{-\frac{2\pi i \cdot 0}{99}})(z - e^{-\frac{2\pi i \cdot 1}{99}})(z - e^{-\frac{2\pi i \cdot 2}{99}}) \cdots (z - e^{-\frac{2\pi i \cdot 98}{99}}). \]
This simplifies to:
\[ P(z) = z^{99} - 1. \]
Next, we need to compute \( P(-2) \):
\[ P(-2) = (-2)^{99} - 1. \]
Calculating \( (-2)^{99} \):
\[ (-2)^{99} = -2^{99}. \]
Thus:
\[ P(-2) = -2^{99} - 1. \]
The final result, therefore, is:
\[ \boxed{-2^{99} - 1}. \]