An organic compound "P" of molecular formula $$C_6H_{12}O_3$$ gives positive Iodoform test but negative Tollen's test. When "P" is treated with dilute acid, it produces "Q". "Q" gives positive Tollen's test and also iodoform test. The structure of "P" is :

Option A: 

Structure: $$CH₃-CO-CH(OCH₃)-CH₂(OCH₃)$$

Although it contains a carbonyl group, the arrangement of $$-OCH₃$$ groups does not form a proper acetal. On hydrolysis, it does not produce an aldehyde ($$-CHO$$) because both $$OCH₃$$ are not on same carbon Therefore, the product formed will not give a positive Tollens’ test. This contradicts the condition given for compound Q.

Hence, Option (A) is not correct.

Option B:

Structure of P: $$CH₃-CO-CH₂-CH(OCH₃)₂$$

Iodoform test is given by compounds containing the $$CH₃-CO-$$ group (methyl ketone) or $$CH₃-CH(OH)-$$ group.
Tollens’ test is given by aldehydes ($$-CHO$$ group).

In compound P, the $$CH₃-CO-$$ group is present, so it gives a positive iodoform test. There is no $$-CHO$$ group, so Tollens’ test is negative. The $$-CH(OCH₃)₂$$ group is an acetal, which on hydrolysis in dilute acid converts into an aldehyde.

On hydrolysis:
$$CH₃-CO-CH₂-CH(OCH₃)₂$$ → $$CH₃-CO-CH₂-CHO$$

Compound Q formed contains both $$-CHO$$ and $$CH₃-CO-$$ groups. Hence, it gives a positive Tollens’ test (due to $$-CHO$$) and a positive iodoform test (due to $$CH₃-CO-$$ group).

Therefore, Option (B) satisfies all the given conditions.

Option C:

Structure: $$H-CO-CH₂-CH₂-CH(OCH₃)₂$$

This compound already contains an aldehyde group ($$-CHO$$). Therefore, it should give a positive Tollens’ test initially. However, the question states that compound P gives a negative Tollens’ test. This is a contradiction.

Hence, Option (C) is not correct.

Option D:

Structure: $$CH₃-CO-C(OCH₃)₂-CH₃$$

This compound is a ketal. On hydrolysis, ketals give ketones and not aldehydes. Therefore, the product Q will not contain a $$-CHO$$ group and will not give a positive Tollens’ test. This does not match the given condition.

Hence, Option (D) is not correct.

Thus, option B is the correct choice.