In proportional allocation the stratum sample sizes $n_h=nW_h$ are fixed by design. In post-stratification the units are classified into strata only after the sample is drawn, so the realised $n_h$ are random — multinomial rather than fixed.

That extra randomness costs precision. Conditioning on the realised counts and then averaging gives, to order $n^{-2}$ ,

$V\bigl(\overline{y}_{\text{post}}\bigr) \approx \underbrace{\frac{1}{n}\sum_{h}W_hS_h^{2}}_{=\,V(\overline{y}_{st})_{\text{prop}}} \;+\; \underbrace{\frac{1}{n^{2}}\sum_{h}(1-W_h)S_h^{2}}_{\text{price of random } n_h}\;,$

the second term being strictly positive. Hence

$V\bigl(\overline{Y}_{st}\bigr)_p < V\bigl(\overline{Y}_{st,\text{post}}\bigr),$

which is option (B). Post-stratification is still typically far better than plain SRS — it recovers most of the gain from stratification when the stratum membership is unknown at the design stage — but it cannot beat genuine proportional allocation.

Option (A) reverses the inequality. Option (C) is only an asymptotic statement (the penalty is $O(n^{-2})$ against a leading term of $O(n^{-1})$ ), not the exact relation asked for. Option (D) is not the post-stratified variance formula. The board's key marks the question defective — presumably because (C) is also defensible to first order — but (B) is the exact relation.