Answer: 6

Solution:

Polynomial division gives $$\begin{aligned} &P(x) = Q(x)\,(x^2 + 1) \\ &\quad {}+ x^2 - x + 1, \end{aligned}$$ so $P(z_i) = z_i^2 - z_i + 1$ for each root $z_i$ of $Q.$

By Vieta's formulas for $Q(x) = x^4 - x^3 - x^2 - 1,$ we have $\sum z_i = 1$ and $\sum_{i \lt j} z_i z_j = -1,$ so $\sum z_i^2 = \left(\sum z_i\right)^2 - 2\sum_{i \lt j} z_i z_j$ $= 1 + 2 = 3.$ Therefore $$\sum_{i=1}^4 P(z_i) = 3 - 1 + 4 = 6.$$