= Solution
Put $s=1-p$. On the four-cycle, the total unnormalized weight is
$$
Z=p^4q+4p^3sq+6p^2s^2q^2+4ps^3q^3+s^4q^4.
$$
The event $N\leftrightarrow S$ contains all configurations with zero or one closed edge and exactly two of the six configurations with two closed edges, so its weight is
$$
p^4q+4p^3sq+2p^2s^2q^2.
$$
Divide numerator and denominator by $q$. As $s\to0$ and $sq\to\infty$, the omitted numerator terms are $o(1+2s^2q)$, while the middle denominator terms are
$$
o(1+s^4q^3).
$$
Consequently
$$
\mathbb P_{\mathrm{FK}}^{p,q}(N\leftrightarrow S)
=(1+o(1))
\frac{2(1-p)^2q+1}{(1-p)^4q^3+1}.
$$
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