= Solution
Set the time derivatives in the <master equation> to zero. The first and third rows give $f_1p_0=f_3p_B$ and $f_2p_A=(f_3+f_4)p_B$. Normalization then gives, with $\mathcal N=f_2f_3+f_1(f_3+f_4)+f_1f_2$,
$$
\boxed{p_0^{\rm st}=\frac{f_2f_3}{\mathcal N},\qquad p_A^{\rm st}=\frac{f_1(f_3+f_4)}{\mathcal N},\qquad p_B^{\rm st}=\frac{f_1f_2}{\mathcal N}.}
$$
These expressions assume positive rates, or a limiting choice for which $\mathcal N\ne0$ 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.
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