I want to show that $$P(X\!=\!i, Y=j) = \frac{p(i \mid j)}{\displaystyle \sum_i \frac{p(i \mid j)}{q(j \mid i)}}~,$$ where $p(i \mid j) = P(X\!=\!i \mid Y\!=\!j)$ and $q(j \mid i) = P(Y\!=\!j \mid X\!=\!i)$.
My attempt:
$$P(X\!=\!i, Y\!=\!j) = P(Y\!=\!j)\,P(X\!=\!i \mid Y\!=\!j) = P(Y\!=\!j) \,p(i \mid j)~.$$
If we can show that $P(Y\!=\!j) = \left( \sum_i \frac{p(i|j)}{q(j|i)} \right)^{-1}$, we are done.
How can we show that? I got $$P(Y\!=\!j) = \sum_i P(X\!=\!i, Y\!=\!j) = \sum_i P(Y\!=\!j)\, P(X\!=\!i \mid Y\!=\!j) = {}??$$
I suppose that $q(j|i)=P(Y=j|X=i)$ (in order not to use $p$ for two different things). Observe that for any $i$ and $j$ in the meaningful domains \begin{align*} \frac{p(i|j)}{q(j|i)} &= \frac{P(X=i|Y=j)}{P(Y=j|X=i)}\\ &= \frac{\frac{P(X=i,Y=j)}{P(Y=j)}}{\frac{P(X=i,Y=j)}{P(X=i)}}\\ &= \frac{P(X=i)}{P(Y=j)} \end{align*} Hence if we sum over i, we get \begin{align*} \sum_i \frac{p(i|j)}{q(j|i)} &= \sum_i\frac{P(X=i)}{P(Y=j)}\\ &= \frac{1}{P(Y=j)} \sum_i P(X=i)\\ &= \frac{1}{P(Y=j)} \end{align*}