A complex number needs a complex problem.

Since $|z|=1$ , $\implies z = e^{i\theta}$ , where $\theta$ is the argument of $z$ . Then

$\begin{aligned} P & = |z+1|+|z^2-z+1| \\ & = |e^{i\theta} +1 | + |e^{2i\theta} - e^{i\theta} + 1| \quad \quad \small \color{#3D99F6} \text{By Euler's formula: }e^{i\theta} = \cos \theta + i\sin \theta \\ & = |\cos \theta + i\sin \theta +1 | + |\cos 2\theta + i\sin 2\theta - \cos \theta - i\sin \theta + 1| \\ & = \sqrt{(\cos \theta +1)^2 + \sin^2 \theta} + \sqrt{(\cos 2\theta - \cos \theta +1)^2 + (\sin 2\theta - \sin \theta)^2} \\ & = \sqrt{\cos^2 \theta + 2\cos \theta +1 + \sin^2 \theta} + \sqrt{(2\cos^2 \theta - 1 - \cos \theta +1)^2 + (2\sin \theta \cos \theta - \sin \theta)^2} \\ & = \sqrt{2 + 2\cos \theta} + \sqrt{\cos^2 \theta(2\cos \theta - 1)^2 + \sin^2 \theta (2\cos \theta - 1)^2} \\ & = \sqrt{2 + 4\cos^2 \frac \theta 2-2} +|2\cos \theta - 1| \\ & = \left|2\cos \frac \theta 2\right| +|2\cos \theta - 1| \end{aligned}$

We note that $P(\theta)$ has a period of $2\pi$ and is symmetrical about $\pi$ . We only need to consider for the domain of $0 \le \theta \le \pi$ .

For $0 \le \theta < \frac \pi 3$ , $P = 2\cos \frac \theta 2 +2\cos \theta - 1$ , which is a decreasing function, $\implies \sqrt 3 < P \le 3$ .

For $\frac \pi 3 \le \theta \le \pi$ ,

$\begin{aligned} P & = 2 \cos \frac \theta 2 + 1 - 2\cos \theta \\ & = 2 \cos \frac \theta 2 + 1 - 4\cos^2 \frac \theta 2 + 2 \\ & = \frac {13}4 - 4\left(\cos \frac \theta 2 - \frac 14 \right)^2 \end{aligned}$

Therefore, $\begin{cases} \max(P) = \frac {13}4 - 4\min \left(\cos \frac \theta 2 - \frac 14 \right)^2 = \frac {13}4 & \text{when } \theta = \cos^{-1} \frac 14 \in \left[\frac \pi 3, \pi \right] \\ \min(P) = \frac {13}4 - 4\max \left(\cos \frac \theta 2 - \frac 14 \right)^2 = \sqrt 3 & \text{when } \theta = \frac \pi 3 \end{cases}$

$\implies P_{\max} = \frac {13}4$ , $P_{\min} = \sqrt 3$ , and $P_{\max}P_{\min} = \boxed{\frac {13\sqrt 3}4}$ .