The problem describes a sequence of organic reactions starting from a tertiary alcohol H. We need to identify the structures of intermediate compounds J, K, and L.

Step-by-step Derivations:

1. Analyze the reaction of Compound J:

   • The problem states that compound J reacts with $KOH$ to give benzyl alcohol ($PhCH_2OH$) and a compound L.
   • The reaction of a carbonyl compound with a strong base ($KOH$) to produce an alcohol and a carboxylate salt is characteristic of the Cannizzaro reaction.
   • The Cannizzaro reaction occurs with aldehydes that lack α-hydrogens.
   • Since one of the products is benzyl alcohol (the reduction product), the starting compound J must be benzaldehyde ($PhCHO$). Benzaldehyde has no α-hydrogens.
   • The reaction is a disproportionation where one molecule of benzaldehyde is reduced to benzyl alcohol and the other is oxidized to benzoic acid, which forms a salt with $KOH$. $2 \, PhCHO \xrightarrow{KOH} PhCH_2OH + PhCOO^-K^+$
   • From this, we can identify:
     • J = Benzaldehyde ($PhCHO$)
     • L = Potassium benzoate ($PhCOO^-K^+$)

2. Evaluate the Options based on J and L:

   • Based on our identification of J as $PhCHO$ and L as $PhCOO^-K^+$, we can examine the given options:
     • A: J = $PhCOCH_3$ (incorrect)
     • B: J = $PhCHO$, L = $PhCOO^-K^+$ (possible)
     • C: J = $PhCOCH_3$ (incorrect)
     • D: J = $PhCHO$, L = $PhCOO^-K^+$ (possible)
   • We have narrowed down the correct answer to either B or D.

3. Determine the structure of Compound K and Alkene I:

   • Compounds J and K are formed from the ozonolysis of an alkene I. Ozonolysis cleaves the double bond of the alkene and forms two carbonyl compounds.
   • The alkene I can be reconstructed by joining the carbonyl carbons of J and K with a double bond.
   • Let's consider the two possibilities for K from options B and D:
     • If K is $PhCH_2CHO$ (from option B): The alkene I would be formed from $PhCHO$ (J) and $PhCH_2CHO$ (K). $I = Ph-CH=CH-CH_2-Ph \quad \text{(1,3-diphenylpropene)}$
     • If K is $PhCOCH_3$ (from option D): The alkene I would be formed from $PhCHO$ (J) and $PhCOCH_3$ (K). $I = Ph-CH=C(CH_3)Ph \quad \text{(1,2-diphenylpropene)}$

4. Trace back to the Tertiary Alcohol H:

   • Alkene I is formed by the acid-catalyzed dehydration of a tertiary alcohol H.
   • Let's check which of the possible structures for I could be formed from a tertiary alcohol.
     • For I = 1,3-diphenylpropene ($Ph-CH=CH-CH_2-Ph$): The precursor alcohol would have an -OH group on a carbon atom adjacent to the double bond. Possible structures are $Ph-CH_2-CH(OH)-CH_2-Ph$ or $Ph-CH(OH)-CH_2-CH_2-Ph$. Both of these are secondary alcohols. This contradicts the problem statement that H is a tertiary alcohol. So, option B is incorrect.
     • For I = 1,2-diphenylpropene ($Ph-CH=C(CH_3)Ph$): The precursor alcohol H could be $Ph-CH_2-C(OH)(CH_3)Ph$. This alcohol (2,3-diphenyl-2-butanol) is a tertiary alcohol. Dehydration of this alcohol would proceed via a stable tertiary carbocation, $Ph-CH_2-C^+(CH_3)Ph$. Elimination of a proton from the adjacent $CH_2$ group leads to the formation of the more substituted and conjugated alkene I, which is the major product (Saytzeff's rule). $Ph-CH_2-C(OH)(CH_3)Ph \xrightarrow{H^+} Ph-CH=C(CH_3)Ph + H_2O$
     • This pathway is consistent with all the given information.

5. Final Confirmation:

   • The sequence is as follows:
     • H: $Ph-CH_2-C(OH)(CH_3)Ph$ (tertiary alcohol)
     • I: $Ph-CH=C(CH_3)Ph$
     • Ozonolysis of I gives J and K: $PhCHO$ (J) and $PhCOCH_3$ (K).
     • J ($PhCHO$) + KOH $\rightarrow$ Benzyl alcohol + $PhCOO^-K^+$ (L).
     • K ($PhCOCH_3$) + KOH $\rightarrow$ M. Acetophenone has α-hydrogens and would undergo aldol condensation, which is consistent with the reaction description.

Conclusion:

The structures are:

• J: Benzaldehyde ($PhCHO$)
• K: Acetophenone ($PhCOCH_3$)
• L: Potassium benzoate ($PhCOO^-K^+$)

This corresponds to option D.