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Markov chain with coin flips, 1 success, rest failures.

So I’m trying to understand what the next state is for a Markov chain, where each transition is independent, for a simple $2 \times 2$ matrix $P$. I believe this is relatively straightforward.
$A=\begin{matrix}0.2 & 0.8 \\0.8 & 0.2\end{matrix}$
Let $B=A+I$. Then, $B=\begin{matrix}0.8 & 1 \\1 & 0.8\end{matrix}$
But I’m having trouble finding $P_2$. In particular, I know that $P_2$ is not a function of $P_{11}$ and $P_{12}$. So for instance, $P_{11}=0$ is in a 1 state. But $P_{11}=0$ does not make it impossible for $B$ to make it to a $2$ state. What is $P_{12}$ in this case?
My other thought was that the only states that are not strictly increasing in the number of flips are $2$ and $3$, which both need one success. Is that correct?

It’s easy to see that there’s just a single equivalence class of $0-1$ states.
In your case, suppose $(p_{11},p_{12})$ is in the equivalence class for some $0\leq p_{12}\leq 1-p_{12}\leq 1$.
Consider the $2$-steps

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