Let $(X_n)$ be a Markov chain and ${\cal F_n}=\sigma(X_0,\ldots,X_n)$ the past $\sigma$-field at time $n$. The quite complicated statement of your problem says nothing but that the equality $$\Pr (X_{n+1} = a \mid {\cal F_n})=\Pr(X_{n+1} = a \mid X_n=b)$$ holds on the event $\{X_n=b\}$.

Let $(X_n)$ be a Markov chain and ${\cal F_n}=\sigma(X_{n-1},X_{n-2},\ldots)$ the past $\sigma$-field at time $n$. The quite complicated statement of your problem says nothing but that the equality $$\Pr (X_{n+1} = a \mid {\cal F_n})=\Pr(X_{n+1} = a \mid X_n=b)$$ holds on the event $\{X_n=b\}$.