= Solution
Use the high-temperature state consistently to first order in $\epsilon$. Applying the lower-target <controlled-transfer gate> with $\cos\theta_1=1/3$ gives
$$
\rho_1=\frac14\operatorname{diag}\left(1+\epsilon,1,1-\frac\epsilon3,1-\frac{2\epsilon}3\right).
$$
The upper-target gate with $\cos\theta_2=0$ then equalizes the second and fourth populations. The result is
$$
\boxed{\rho_{\mathrm{pp}}=\frac14\operatorname{diag}\left(1+\epsilon,1-\frac\epsilon3,1-\frac\epsilon3,1-\frac\epsilon3\right)
=(1-p)\frac I4+p|00\rangle\langle00|,\qquad p=\frac\epsilon3.}
$$
This is a <pseudo-pure state>; the identity part is unchanged by unitary gates and does not contribute to traceless <NMR> signals.
For $n$ equal, weakly polarized <spins>, the high-temperature expansion is $\rho\simeq2^{-n}[I+(\epsilon/2)\sum_j Z_j]$. The all-zero population is $a_0=2^{-n}(1+n\epsilon/2)$. Preserving this largest population and averaging the other $2^n-1$ populations gives
$$
\boxed{p_n=\frac{2^na_0-1}{2^n-1}\simeq\frac{n\epsilon}{2(2^n-1)}.}
$$
Thus the signal-bearing effective purity falls approximately as $n/2^n$ at fixed small per-spin <polarization>. The linear expansion requires $n|\epsilon|\ll1$. More generally, if the exact single-spin <polarization> is $r=\tanh(\epsilon/2)$, the same population-preserving averaging gives $p_n=[(1+r)^n-1]/(2^n-1)$, which still decreases exponentially for fixed $0<r<1$. This scaling concerns high-temperature ensemble preparation, not an impossibility theorem for every initialization method.
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