jmc

The first two vertices of $V$ have magnitude $\sqrt{2}$, while the other four have magnitude $\frac{1}{2}$. In order for $P=-1$, it must be the case that $|P|=1$, which only happens if there are two magnitude-$\sqrt{2}$ vertices for each magnitude-$\frac{1}{2}$ one. Define $P_1$ as the product of the magnitude-$\sqrt{2}$ vertices chosen and $P_2$ as the product of the magnitude-$\frac{1}{2}$ vertices chosen.

There are $\left(\genfrac{}{}{0ex}{}{12}{8}\right)$ ways to select which of the 12 draws come up with a magnitude-$\sqrt{2}$ number. The arguments of those numbers are all $\pm \frac{\pi }{2}$, so $P_1$ has an argument that is a multiple of $\pi$. Half of the $2^8$ draw sequences will produce a result with argument equivalent to $0$ and the other half will have an argument equivalent to $\pi$.

Similarly, the arguments of the other four numbers are $\frac{\pi }{4}+k\cdot \frac{\pi }{2}$, so $P_2$ has argument $k\cdot \frac{\pi }{2}$ for some integer $k$. The $4^4$ ways to select four magnitude-$\frac{1}{2}$ numbers are equally likely to produce any of the four possible product arguments.

In order for $P=-1$, the argument of the product must be $-\frac{\pi }{2}$. That happens only if: (a) $P_1$ has argument $0$ and $P_2$ has argument $-\frac{\pi }{2}$, which happens with probability $\frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$. (b) $P_2$ has argument $\pi$ and $P_2$ has argument $\frac{\pi }{2}$, which also happens with probability $\frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$. Putting these cases together, we find that $\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$ of the $2^8\cdot 4^4=2^{16}$ sequences of eight magnitude-$\sqrt{2}$ and four magnitude-$\frac{1}{2}$ vertices will have the correct argument for $P=-1$.

The probability that $P=-1$ is $$\begin{array}{rl}\frac{\left(\genfrac{}{}{0ex}{}{12}{4}\right)\cdot \frac{1}{4}\cdot 2^{16}}{6^{12}}& =\frac{\left(\genfrac{}{}{0ex}{}{12}{4}\right)4}{3^{12}}\\ & =\frac{12\cdot 11\cdot 10\cdot 9\cdot 4}{4!\cdot 3^{12}}\\ & =\frac{220}{3^{10}}.\end{array}$$The final answer is $220+3+10=\boxed{233}.$