The major product of the following reaction is:

1. Reduction of the Nitrile Group ($$-\text{CN}$$):

   $$\text{DIBAL-H}$$ partially reduces the nitrile group attached to the aromatic ring into an imine intermediate. Upon subsequent acid hydrolysis ($$\text{H}_3\text{O}^\oplus$$), this imine is converted directly into a formyl group (aldehyde, $$-\text{CH}=\text{O}$$):

   $$\text{Ar--C}\equiv\text{N} \xrightarrow{\text{(i) DIBAL-H}} \text{Ar--CH}=\text{NH} \xrightarrow{\text{(ii) H}_3\text{O}^\oplus} \text{Ar--CH}=\text{O}$$



2. Reduction and Cleavage of the Lactone (Cyclic Ester):

   $$\text{DIBAL-H}$$ also attacks the carbonyl carbon of the cyclic lactone ring, reducing it to a cyclic hemiacetal stage. During the acidic workup ($$\text{H}_3\text{O}^\oplus$$), this ring opens up completely to release a phenolic hydroxyl group ($$-\text{OH}$$) on the aromatic ring and a terminal aliphatic aldehyde group ($$-\text{CH}_2\text{CH}=\text{O}$$) on the side chain.

Conclusion:

Both the nitrile and lactone functional groups are cleanly reduced by $$\text{DIBAL-H}$$ to yield a product containing a phenol and two distinct aldehyde functional groups.

Answer: Option A