We match each pair in List I with the correct isomerism type in List II.

Step 1: Identify each pair

A. Propanamine and N-Methylethanamine

• Propanamine: $\mathrm{CH_3CH_2CH_2NH_2}$
• N-Methylethanamine: $\mathrm{CH_3CH_2NHCH_3}$

Both have the same molecular formula $\mathrm{C_3H_9N}$, but belong to different classes of amines:

• one is a primary amine
• the other is a secondary amine

Thus, they are functional isomers.

So, $A \to III$

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B. Hexan-2-one and Hexan-3-one

• Hexan-2-one: carbonyl at position 2
• Hexan-3-one: carbonyl at position 3

Same carbon skeleton and same functional group, but different position of the ketone group.

Thus, they are positional isomers.

So, $B \to II$

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C. Ethanamide and Hydroxyethanimine

• Ethanamide: $\mathrm{CH_3CONH_2}$
• Hydroxyethanimine: $\mathrm{CH_3C(OH)=NH}$

These two are related by proton shift and movement of double bond: $\mathrm{CH_3CONH_2 \rightleftharpoons CH_3C(OH)=NH}$

This is amide-imidic acid tautomerism.

Thus, they are tautomers.

So, $C \to IV$

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D. $o$-Nitrophenol and $p$-Nitrophenol

These differ in the position of the nitro group relative to the hydroxyl group on benzene ring:

• ortho position
• para position

Thus, they are positional isomers.

So, $D \to II$

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Step 2: Compare with given options

From the above:

• $A \to III$
• $B \to II$
• $C \to IV$
• $D \to II$

Now check the options:

• Option A: $A$-$III$, $B$-$I$, $C$-$IV$, $D$-$II$
  Here $B$ is incorrect.
• Option B: $A$-$III$, $B$-$IV$, $C$-$I$, $D$-$II$
  Incorrect.
• Option C: $A$-$II$, $B$-$III$, $C$-$I$, $D$-$IV$
  Incorrect.
• Option D: $A$-$IV$, $B$-$III$, $C$-$I$, $D$-$II$
  Incorrect.

There is no option matching the chemically correct classification exactly.

However, in standard exam convention, Hexan-2-one and Hexan-3-one are definitely positional isomers, not metamers. So the stored answer appears inconsistent with chemistry.

The nearest intended matching is likely: $A\to III,\; B\to II,\; C\to IV,\; D\to II$ which is not present in the options.