 Solution

Set the time derivatives in the master equation to zero. The first and third rows give $f_1{p}_0=f_3{p}_B$ and $f_2{p}_A=(f_3+f_4)p_B$. Normalization then gives, with $\mathcal{N}=f_2{f}_3+f_1(f_3+f_4)+f_1{f}_2$,These expressions assume positive rates, or a limiting choice for which $\mathcal{N}\ne 0$ and the stationary distribution is unique. If rates vanish so that multiple closed classes appear, the initial condition can matter.

The word “equilibrium” in this part must mean a stationary distribution. With positive $f_1,f_2,f_3$ the one-way contact transitions support a nonzero stationary electric current, so the process is not in thermodynamic equilibrium and does not satisfy detailed balance. In particular, $0\to A$ carries positive probability flow while the direct reverse rate is zero.