Step 1: Identify Isomers P and Q

1. Analyze the molecular formula and properties: The formula is $C_4H_4O_4$. It is a dicarboxylic acid. Let's calculate the Degree of Unsaturation (DoU). $DoU = C + 1 - \frac{H}{2} = 4 + 1 - \frac{4}{2} = 3$
2. Account for functional groups: A dicarboxylic acid has two carboxyl groups (-COOH). Each -COOH group contains a C=O double bond, accounting for one degree of unsaturation. So, two -COOH groups account for 2 degrees of unsaturation.
3. Determine the remaining unsaturation: The total DoU is 3, and the carboxyl groups account for 2. The remaining degree of unsaturation (3 - 2 = 1) must be in the carbon chain. The fact that both isomers decolorize $Br_2/H_2O$ indicates the presence of a carbon-carbon double bond (C=C).
4. Deduce the structures: The structure is a 4-carbon chain with a double bond and two carboxyl groups: HOOC-CH=CH-COOH (Butenedioic acid). This can exist as two geometric isomers:
   • cis-isomer (Maleic acid): The two -COOH groups are on the same side of the double bond.
   • trans-isomer (Fumaric acid): The two -COOH groups are on opposite sides of the double bond.
5. Identify P and Q: The problem states that "On heating, P forms the cyclic anhydride." Cis-dicarboxylic acids, where the carboxyl groups are close to each other, readily form cyclic anhydrides upon heating by eliminating a water molecule. Trans-isomers do not, as the groups are too far apart.
   • Therefore, P is Maleic acid (cis-isomer).
   • Q is Fumaric acid (trans-isomer).

Step 2: Analyze the reaction sequence starting with P

The reaction sequence is: $P \xrightarrow{\Delta} \xrightarrow{R(C_2H_6O)} V \xrightarrow{1. LiAlH_4, 2. H_3O^+} W$

1. First Reaction (Heating P): P is maleic acid. Heating maleic acid causes dehydration to form maleic anhydride, a cyclic anhydride. $\text{Maleic acid} \xrightarrow{\Delta} \text{Maleic anhydride} + H_2O$

2. Second Reaction (Reaction with R): The product, maleic anhydride, reacts with R ($C_2H_6O$). $C_2H_6O$ is ethanol ($CH_3CH_2OH$). Alcohols react with anhydrides in a nucleophilic acyl substitution reaction to open the ring and form a monoester. $\text{Maleic anhydride} + CH_3CH_2OH \rightarrow \text{Ethyl hydrogen maleate (V)}$ The product V is a monoester of maleic acid. The cis-stereochemistry of the double bond is retained. The structure of V is:

3. Third Reaction (Reduction of V): Compound V is treated with lithium aluminum hydride ($LiAlH_4$), a strong reducing agent, followed by acidic workup ($H_3O^+$).

   • $LiAlH_4$ reduces both carboxylic acids (-COOH) and esters (-COOR) to primary alcohols (-CH$_2$OH).
   • Importantly, $LiAlH_4$ also reduces carbon-carbon double bonds that are conjugated with a carbonyl group (as in $\alpha,\beta$-unsaturated acids/esters).
   • V (ethyl hydrogen maleate) has two carbonyl groups conjugated with the C=C double bond. Therefore, $LiAlH_4$ will reduce both carbonyl groups and the double bond.
     • -COOH is reduced to -CH$_2$OH.
     • -COOCH$_2$CH$_3$ is reduced to -CH$_2$OH (and ethanol is formed as a byproduct).
     • -CH=CH- is reduced to -CH$_2$-CH$_2$-.
   • The overall transformation is: $\text{HOOC-CH=CH-COOC}_2\text{H}_5 \xrightarrow{1. LiAlH_4, 2. H_3O^+} \text{HOCH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{OH}$
   • The final product W is butane-1,4-diol.

Step 3: Match V and W with the given options

• V: Ethyl hydrogen maleate (cis-monoester).
• W: Butane-1,4-diol.

Let's examine the options:

• Option A: Shows V as ethyl hydrogen maleate and W as butane-1,4-diol. This matches our derived structures.
• Option B: Shows V as diethyl maleate (diester). This is incorrect.
• Option C: Shows V as ethyl hydrogen fumarate (trans-isomer). This is incorrect as the starting material is cis.
• Option D: Shows V as diethyl fumarate (trans diester). This is incorrect.

Therefore, option A is the correct answer.