= Solution
Put $h_0=1$ and $h_r=0$. The <Markov property> and first-step conditioning give $h_i=\sum_jP_{ij}h_j$. Consider a finite history ending at state $i$ and having positive conditional probability given $A$. The chance of eventual absorption at zero after that history is $h_i$: the future depends only on the present state, and if zero has already been reached, absorption keeps the chain there. If the next state is $j$, this chance is $h_j$. Bayes' rule therefore gives
$$
\mathbb P(X_{n+1}=j\mid X_0,\ldots,X_n=i,A)
=\frac{P_{ij}h_j}{h_i}.
$$
This depends on the history only through $i$, proving the conditional <Markov property>. \b[The conditional <transition matrix> is]
$$
\boxed{Q_{ij}=P_{ij}\frac{h_j}{h_i}\quad(0\leq i<r).}
$$
Its row sums are one by harmonicity of $h$, and $Q_{ir}=0$. State $r$ is never visited under the conditioning and is naturally removed; any row assigned there on an enlarged formal state space is immaterial, since conditioning starting at $r$ is undefined. State zero stays absorbing. For a random initial law $\nu$, its conditional law is $\nu_i h_i/\sum_j\nu_jh_j$. This is the <Doob h-transform> with the absorption probability as harmonic function.
For the simple symmetric walk, $h_i=(r-i)/r$ solves the linear harmonic recurrence and its boundary values. Thus
$$
Q_{i,i-1}=\frac{r-i+1}{2(r-i)},\qquad
Q_{i,i+1}=\frac{r-i-1}{2(r-i)},\qquad 1\leq i<r.
$$
In particular, the conditioned chain moves downward with probability one from $r-1$.
Let $\tau$ be the first hit of $\{0,r\}$ in the original walk and let $g_i=\mathbb E_i[\tau\mathbf1_A]$. Its absorption time has finite expectation: from any transient state a run of at most $r$ downward steps hits zero with probability at least $2^{-r}$, giving a geometric tail bound by blocks. First-step conditioning then gives
$$
g_0=g_r=0,\qquad g_i=h_i+\frac12(g_{i-1}+g_{i+1}).
$$
Equivalently, $g_{i+1}-2g_i+g_{i-1}=-2(r-i)/r$. The cubic <polynomial>
$$
g_i=\frac{i(r-i)(2r-i)}{3r}
$$
has exactly these second differences and boundary values. It is the unique solution, since the difference of any two solutions is a linear sequence vanishing at both endpoints. Conditional on $A$, $\tau$ is the first time in zero, so division by $h_i$ proves
$$
\boxed{\mathbb E_i[\tau_0\mid A]=\frac{g_i}{h_i}=\frac{i(2r-i)}3.}
$$
This is the <duration of gambler's ruin conditioned on a chosen boundary>.
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