Let $r$ be the number of countries in Wonderland. If the minimal number of women in working groups is $n = 0$ then forming a working group means just choosing one government member from each country. Thus there are $(a+b)^r$ different working groups. This number can be prime only if $r = 1$ because $a+b \ge b > 1$.

If the minimal number of women in working groups is $n \ge 1$ then a working group containing exactly $k$ women ($n \le k \le r$) can be formed as follows. Choose $k$ countries out of $r$, that send a woman to that particular working group, then choose one woman out of $b$ from each of the $k$ governments, and finally choose one man out of $a$ from each of the remaining $r-k$ countries. Hence there are $\binom{r}{k} b^k a^{r-k}$ working groups with exactly $k$ women, and $\sum_{k=n}^{r} \binom{r}{k} b^k a^{r-k}$ working groups with at least $n$ women altogether. As $n \ge 1$, all terms of this sum are divisible by $b$, whence the sum can be a prime only if it is equal to $b$. This is possible only if $r=1$ since otherwise the last term (corresponding to $k=r$) of the sum would be greater than $b$.