Consider $$P = \begin{pmatrix} 0 & a & 0 & 1-a\\ 1-b & 0 & b & 0\\ 0 & 1-c & 0 & c\\ d & 0 & 1-d & 0\\ \end{pmatrix}$$ with $S = \{1,2,3,4\}$, and show that there is a stationary distribution satisfying the detailed balance condition if $$0 < abcd = (1 - a)(1 - b)(1 - c)(1 - d).$$
I know that the detailed balance condition requires $\pi_i p(i,j) = \pi_j p(j,i)$ for all $i, j \in S$. Thus, my initial idea was to rewrite the equality as follows: $$abcd = p(1,2) p(2,3) p(3,4) p(4,1)$$ and $$(1 - a)(1 - b)(1 - c)(1 - d) = p(2,1) p(3,2) p(4,3) p(1,4)$$ However, I'm unsure where to go from this point. I tried thinking about properties of the stationary distribution ($\sum_{i \in S} \pi_i = 1$, $\pi_2 p(1,2) + \pi_3 p(1,3) + \pi_4 p(1,4) = \pi_1$), but have thus far not been able to transform this into the detailed balance condition. Any advice is greatly appreciated!
We can think of the four states of this Markov chain as being arranged in a cycle. We go clockwise from $i$ to $(i+1)\bmod 4$ with probabilities $a$, $b$, $c$, $d$ respectively, and counterclockwise from $i$ to $(i-1) \bmod 4$ with probabilities $1-a$, $1-b$, $1-c$, $1-d$, respectively.
Any time the detailed balance conditions are satisfied, if we start at one value of $\pi_i$, we can deduce the values of all others connected to it. In this example, we can write down $\pi_2$, $\pi_3$, and $\pi_4$ in terms of $\pi_1$: \begin{align} \pi_1 a = \pi_2 (1-b) &\implies \pi_2 = \frac{a}{1-b} \pi_1 \\ \pi_2 b = \pi_3 (1-c) &\implies \pi_3 = \frac{b}{1-c} \pi_2 = \frac{ab}{(1-b)(1-c)}\pi_1 \\ \pi_3 c = \pi_4 (1-d) &\implies \pi_4 = \frac{c}{1-d} \pi_3 = \frac{abc}{(1-b)(1-c)(1-d)}\pi_1. \end{align} With just these three equations, which are the detailed balance equations for $(1,2)$, $(2,3)$, and $(3,4)$, we've solved for $\pi_2, \pi_3, \pi_4$ in terms of $\pi_1$.
But there's one more equation we haven't used yet. It says that $\pi_4 d = \pi_1(1-a)$ and since we already know both sides, we can write it as $$\frac{a b c}{(1-b)(1-c)(1-d)}\pi_1 d = \pi_1 (1-a) \iff a b c d = (1-a)(1-b)(1-c)(1-d).$$ (We canceled $\pi_1$ from both sides, but that's okay: if we had $\pi_1 = 0$ then the equations above would tell us $\pi_2 = \pi_3 = \pi_4 = 0$ as well, which wouldn't be any good.)
In other words, starting with an arbitrary $\pi_1$, we can force detailed balance at $(1,2)$, $(2,3)$, and $(3,4)$ to hold by choosing $\pi_2, \pi_3, \pi_4$ appropriately. (We could now also work out $\pi_1$ by using the fact that $\pi_1 + \pi_2 + \pi_3 + \pi_4 = 1$, but we don't need to.) But we no longer have the freedom to force detailed balance to hold at $(1,4)$, since $\pi_1$ and $\pi_4$ are fixed. We can just hope that the equation turns out to hold anyway, and the above calculation shows that it does hold exactly when $a b c d = (1-a)(1-b)(1-c)(1-d)$.
(Once that's true, we have a solution that satisfies detailed balance, and therefore it must also be a stationary distribution.)