= Solution
Work on the single event of <probability> one where $Y_n/n\to\mu=2p-1>0$. On every <sample path> in that event, $Y_n\to+\infty$ and the unit-step property ensures that every positive <integer> $j$ is reached, so $T_j<\infty$. Starting from zero requires at least $j$ steps to reach $j$, hence $T_j\geq j$ and $T_j\to\infty$.
The convergent numerical sequence $Y_n/n$ on this fixed <sample path> also converges along its possibly random subsequence $n=T_j$. Since $Y_{T_j}=j$,
$$
\frac{j}{T_j}=\frac{Y_{T_j}}{T_j}\longrightarrow\mu.
$$
Taking the reciprocal is valid because $\mu>0$. We obtain
$$
\boxed{\mathbb P\!\left(\frac{T_j}{j}\longrightarrow\alpha\right)=1,
\qquad\alpha=\frac1{2p-1}.}
$$
This is a <first-passage law of large numbers for an upward skip-free random walk>. No independence between the <hitting times> and the path is needed: <almost sure convergence> already supplies convergence along every diverging subsequence on the same event.
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