= Solution
A syndrome determines a phase-error set only up to multiplication by $\overline X=Z_1\cdots Z_n$: an error of weight $w$ and its complement of weight $n-w$ have the same <error syndrome>. Their probability ratio is
$$
\frac{p^w(1-p)^{n-w}}{p^{n-w}(1-p)^w}
=\left(\frac p{1-p}\right)^{2w-n}.
$$
For $p<1/2$, <maximum likelihood estimation>[maximum likelihood] therefore chooses the representative of smaller weight. This is exactly <majority-vote decoding of a repetition code>: correction succeeds when fewer than half the qubits are flipped, with a random tie-break at $w=n/2$ when $n$ is even.
If $W\sim\operatorname{Bin}(n,p)$, then $\mathbb EW=np$ and $\operatorname{sd}(W)=\sqrt{np(1-p)}$. For every fixed $p<1/2$, the distance from the mean to the decision boundary in standard deviations is
$$
\frac{n/2-np}{\sqrt{np(1-p)}}
=\sqrt n\,\frac{1/2-p}{\sqrt{p(1-p)}}\longrightarrow\infty.
$$
Thus the logical failure probability $\Pr(W>n/2)$ tends to zero, in fact exponentially by a <Chernoff bound>. At $p=1/2$ the two representatives are equiprobable and decoding cannot improve with $n$. Hence
$$
\boxed{p_c=\frac12.}
$$
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