Step 1: Identify the compound in (D).
Starting with the given compound \(\text{C}_4\text{H}_9\text{Br}\), which is a bromoalkane, the reaction with alcoholic KOH results in elimination, forming an alkene (product A). The next step involves treating this alkene with sodium amide (NaNH\(_2\)) at high temperature, which causes further elimination (product B), resulting in an alkyne. The subsequent reaction with \(\text{HgSO}_4\) and dilute sulfuric acid leads to the formation of a ketone, specifically a methyl ketone. Finally, the treatment with bromine (\(Br_2\)) in carbon tetrachloride (\(CCl_4\)) gives the product in (D), which is a methyl ketone (acetone).
Step 2: Confirming the functional group in (D).
Since the final product in (D) is a methyl ketone, the confirmation test for the presence of a methyl ketone functional group is the Haloform test. This test is specific for compounds containing a methyl group attached to a carbonyl group (methyl ketones). In the presence of halogens, such compounds form a characteristic yellow precipitate of chloroform (CHCl\(_3\)).
Step 3: Conclusion.
Thus, the correct answer is option (A) - the Haloform test will confirm the presence of the functional group in (D).
Final Answer: Option (A) Haloform test.