= Solution
For an irreducible <positive recurrent> <Markov chain>, the <Markov chain ergodic theorem> gives a unique <stationary distribution> $\pi$, with almost-sure long-run occupation fraction $\pi_j$ in state $j$. More generally, time averages of an integrable state reward converge to its stationary mean. If the chain is also aperiodic, then $\mathbb P_i(X_n=j)\to\pi_j$ for every initial state $i$; the aperiodicity assumption is needed for this last distributional conclusion.
For either transition rule in part (i), $0<p<1$ permits progression through the phase-A states and a return from phase B to zero. Both chains are finite and irreducible, hence <positive recurrent>. State $K$ has a positive self-loop, so both are aperiodic. Therefore
$$
\boxed{\lim_{n\to\infty}\mathbb P(X_n=j)=\pi_j},
$$
using respectively the cumulative or consecutive <stationary distribution> given in part (i), for every starting distribution.
The excluded endpoint cases are simple: at $p=0$, phase B is eventually reached and then absorbing, so the limiting law is the point mass at $K$. At $p=1$, the cumulative rule traps the chain at its current phase-A count (or eventually at zero if started in phase B), so the limit can depend on the initial count; under the consecutive rule zero is eventually absorbing. Irreducibility and the preceding stationary formulas must not be asserted unchanged at these endpoints.
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