If $z_1, z_2, \dots, z_n$ are vertices of $P$ then $|z_1| = |z_2| = \dots = |z_n| = 1$. By the rotation we can assume that $\prod_{i=1}^n z_i = 1$. We know that

$|w-z_1| \cdot |w-z_2| \cdots |w-z_n| = |(w-z_1)\cdots(w-z_n)|$.

Let $P(w) = \prod_{i=1}^n (w-z_i)$ and $P(w) = w^n + Q(w) + 1$, here $Q(w) = a_{n-1}w^{n-1} + \dots + a_1w$.

Consider the $n$th roots of unity $\epsilon_0, \epsilon_1, \dots, \epsilon_{n-1}$. Therefore $\sum_{s=0}^{n-1} \epsilon^s = 0$.

Here $s \not\equiv 0 \pmod{n}$ we get
$$\sum_{i=1}^{n-1} Q(\epsilon_i) = \sum_{j=1}^{n-1} \left( a_j \cdot \sum_{i=0}^{n-1} \epsilon^j \right) = 0$$

a) Assume to the contrary. If there exist $i$ such that $Q(\epsilon_i) = a_i + b_i$, $a_i > 0$ then
$$P(\epsilon_i) = \epsilon_i^n + 1 + Q(\epsilon_i) = 2 + a_i + b_i \cdot i,$$
thus we have $|P(\epsilon_i)| > 2$. Hence $Q(\epsilon_i) = b_i \cdot i$ for arbitrary $i$. Because we have $\sum_{i=0}^{n-1} a_i = 0$ and all of $a_i$ is less than or equal to zero, thus $a_i = 0$ for arbitrary $i$. Now suppose that for arbitrary $i$: $b_i \neq 0$ then
$$P(\epsilon_i) = \epsilon_i^n + 1 + Q(\epsilon_i) = 2 + b_i \cdot i.$$
So $|P(\epsilon_i)| > 2$. This leads contradiction.

Now for arbitrary $i$, we have $Q(\epsilon_i) = 0$, $i = \overline{0, n-1}$. But $\deg Q(w) = n-1$, hence $Q(w)$ is zero polynomial.
Thus $P(w) = w^n + 1$, we get $|P(\epsilon_i)| = |\epsilon_i^n + 1| = 2$

b) From part a), the multiplication greater than or equal to $2$, it must be $P(w) = w^n + 1$, $w^n + 1 = 0$ equation's solutions are
$$z_i = \cos \frac{2\pi}{2n}(2k + 1) + i \sin \frac{2\pi}{2n}(2k + 1), \quad k = \overline{0, n-1}$$