The compound A in the following reactions is:
$$A \xrightarrow[(ii)Conc. H_{2}SO_{4}/ \Delta]{(i)CH_{3}MgBr/H_{2}O} B \xrightarrow[(ii) Zn/H_{2}O]{(i) O_{3}} C + D$$

The sequence starts with the addition of the Grignard reagent $$CH_3MgBr$$ to compound $$A$$ followed by acidic hydrolysis. A Grignard reagent adds the alkyl group ($$CH_3$$ here) to the carbonyl carbon of an aldehyde or ketone and converts the carbonyl (>C=O) into an $$-OH$$ group.

Let us assume $$A$$ is an aldehyde/ketone with the structure $$R{-}CHO$$ or $$R{-}CO{-}R'$$.
Addition of $$CH_3MgBr$$ to such a carbonyl gives an alcohol in which one of the groups attached to the carbon bearing the $$-OH$$ is the newly introduced $$CH_3$$.

Step (ii) on the same intermediate is conc. $$H_2SO_4/\Delta$$. Conc. sulfuric acid at high temperature dehydrates alcohols to alkenes. Hence compound $$B$$ must be an alkene obtained by dehydration of the alcohol formed in step (i).

Ozonolysis of an alkene in the presence of $$Zn/H_2O$$ cleaves the C=C bond at both ends and gives carbonyl fragments (aldehydes/ketones). Therefore, the nature of products $$C$$ and $$D$$ tells us the carbon-skeleton of $$B$$ which in turn reveals the structure of the alcohol and finally of $$A$$.

Try the simplest aldehyde, acetaldehyde $$CH_3CHO$$, as $$A$$:

Case 1: $$A = CH_3CHO$$ (acetaldehyde)
(i) $$CH_3CHO + CH_3MgBr \xrightarrow{H_2O} CH_3CH(OH)CH_3$$ (2-propanol)
(ii) $$CH_3CH(OH)CH_3 \xrightarrow{conc.\;H_2SO_4/\Delta} CH_3CH=CH_2$$ (propene, $$B$$)

Ozonolysis of propene:

$$CH_3CH=CH_2 \xrightarrow{O_3/(Zn,H_2O)} CH_3CHO + HCHO$$

Thus $$C = CH_3CHO$$ (acetaldehyde) and $$D = HCHO$$ (formaldehyde). Two different carbonyl products indeed appear, matching the statement “$$C + D$$”.

The entire sequence is self-consistent, so $$A$$ must be acetaldehyde $$CH_3CHO$$.

Therefore, the correct choice is Option C which represents acetaldehyde.