Answer: 45

Solution:

Adding $1$ to both sides turns the left side into a perfect fourth power: $$\begin{aligned} &(x - 1)^4 = x^4 - 4x^3 + 6x^2 \\ &\quad {}- 4x + 1 = 2006. \end{aligned}$$ So $x - 1$ is a fourth root of $2006:$ the four roots are $x = 1 \pm \sqrt[4]{2006}$ (real) and $x = 1 \pm i\sqrt[4]{2006}$ (nonreal).

The product of the conjugate pair of nonreal roots is $$\begin{aligned} &\left(1 + i\sqrt[4]{2006}\right) \\ &\quad {}\cdot \left(1 - i\sqrt[4]{2006}\right) \\ &= 1 + \sqrt{2006}. \end{aligned}$$ Since $44^2 = 1936$ $\lt 2006$ $\lt 2025 = 45^2,$ we have $45 \lt P \lt 46,$ so $\lfloor P\rfloor = 45.$