Treat $P$ as a quadratic in $z^4$ and factor as $(z^4-1)\big(z^4+(4\sqrt3+7)\big)$

Since $4\sqrt3+7=(\sqrt3+2)^2,$ the eight roots lie on two squares; the shortest edges have length $\sqrt2$

Factoring in $z^4,$ $$\begin{aligned} P(z) &=(z^4-1) \\ &\quad {}\cdot\big(z^4+(4\sqrt3+7)\big). \end{aligned}$$ The first factor gives the roots $1,-1,i,-i.$ Since $4\sqrt3+7=(\sqrt3+2)^2$ and $2(\sqrt3+2)=(\sqrt3+1)^2,$ writing $w=\tfrac12(\sqrt3+1)$ the other four roots are $w(\pm1\pm i).$

The eight roots are symmetric about the origin with $4$-fold symmetry, and every segment joining two of them has length at least $\sqrt2.$ Thus any such polygon has perimeter at least $8\sqrt2,$ and the polygon with vertices $1,$ $w(1+i),$ $i,$ $w(-1+i),$ $-1,$ $w(-1-i),$ $-i,$ $w(1-i)$ achieves it.

Thus, the correct answer is B.