= Solution
The channel is $\rho\mapsto(1-p)\rho+pI/2$, so
$$
P(0)=(1-p)|\langle0|\psi\rangle|^2+p/2.
$$
Without Eve the sifted error is $p/2$. If Eve alone produces sifted error $e$ ($1/4$ for BB84, $1/3$ for six-state), the combined error is $(1-p)e+p/2$, strictly above the noise baseline exactly when $p<1$. Thus with known stationary $p$ and enough test bits, Eve is statistically detectable for $0\le p<1$ in either protocol; at $p=1$ the channel is completely mixed and erases the distinction.
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