= Solution
We use the <stationary weights of a reflected birth-death chain>, also verifying why they decide the question. If a <stationary distribution> $\pi$ exists, its equation at zero is $\pi_0=\pi_1d_1=\pi_0u_0$. At $k\geq1$, stationarity gives
$$
\pi_ku_k-\pi_{k+1}d_{k+1}=\pi_{k-1}u_{k-1}-\pi_kd_k.
$$
Thus the edge flow is constant, and its value at the boundary is zero. Every <stationary distribution> must therefore satisfy <detailed balance> on each edge. Since irreducibility forces $\pi_0>0$, it must be proportional to
$$
w_0=1,\qquad w_k=\prod_{j=0}^{k-1}\frac{u_j}{d_{j+1}}.
$$
Conversely, if $\sum_kw_k<\infty$, normalizing these positive weights gives a <probability distribution> satisfying <detailed balance>, hence a <stationary distribution>. Existence of a <stationary distribution> for an <irreducible Markov chain> on a countable <state space> is equivalent to <positive recurrence>.
The ratio is
$$
\frac{w_{k+1}}{w_k}=\frac{(kp+\theta)(k+1+\theta)}{(k+\theta)(k+1)q}\longrightarrow\frac pq.
$$
For $p<1/2$, the limit is less than one, so the <ratio test> gives summability. For $p\geq1/2$, observe that $u_k=p+\theta q/(k+\theta)>p\geq1/2$, whereas $d_{k+1}<q\leq1/2$. Thus $w_{k+1}/w_k>1$ for every $k$, so the weights cannot be summable. This also resolves $p=1/2$ without making an inconclusive appeal to the limiting ratio. Therefore
$$
\boxed{\text{Positive recurrence holds exactly when }0<p<\tfrac12,\quad\theta>0.}
$$
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