The compound(s) that shows(show) positive haloform test is(are) 
A compound gives a positive haloform test (reaction with $\text{X}_2$/$\text{OH}^-$ or $\text{I}_2$/$\text{NaOH}$, forming a haloform $\text{CHX}_3$) if it possesses one of the following structural units:
We analyze each compound for the presence of the $\text{R-CO-CH}_3$ group:
(A) $p$-Methylacetophenone:
$$\text{Compound (A)} = \text{Ar}-\text{CO}-\text{CH}_3$$
This compound has a methyl ketone ($\text{CO}-\text{CH}_3$) group.
Therefore, Compound (A) shows a positive haloform test.
(B) Benzylidineacetone (4-Phenylbut-3-en-2-one):
$$\text{Compound (B)} = \text{Ar}-\text{CH}=\text{CH}-\text{CO}-\text{CH}_3$$
This compound also has a methyl ketone ($\text{CO}-\text{CH}_3$) group.
Therefore, Compound (B) shows a positive haloform test.
(C) Methyl $p$-Toluate:
$$\text{Compound (C)} = \text{Ar}-\text{CO}-\text{OCH}_3$$
This is an ester. It does not possess the $\text{CO}-\text{CH}_3$ group directly attached to the carbonyl carbon.
Therefore, Compound (C) does not show a positive haloform test.
(D) $p$-Methylstyrene (1-Propenylbenzene):
$$\text{Compound (D)} = \text{Ar}-\text{CH}=\text{CH}-\text{CH}_3$$
This is an alkene. It is neither a methyl ketone nor the required secondary alcohol structure.
Therefore, Compound (D) does not show a positive haloform test.
The compounds that show a positive haloform test are (A) and (B).
The correct answer is (A) and (B).
