Compound A C$$_9$$H$$_{10}$$O shows positive iodoform test. Oxidation of A with KMnO$$_4$$/KOH gives acid B C$$_8$$H$$_6$$O$$_4$$. Anhydride of B is used for the preparation of phenolphthalein. Compound A is:

We are given:

Compound A has molecular formula $$C_9H_{10}O$$ and gives a positive iodoform test.

Oxidation of A with $$KMnO_4/KOH$$ gives acid B with molecular formula $$C_8H_6O_4$$.

The anhydride of B is used for the preparation of phenolphthalein.

Step 1: Identify acid B.

Phenolphthalein is prepared by heating phthalic anhydride with phenol in the presence of a dehydrating agent (like concentrated $$H_2SO_4$$ or $$ZnCl_2$$).

Phthalic anhydride is the anhydride of phthalic acid (benzene-1,2-dicarboxylic acid), which has the molecular formula $$C_8H_6O_4$$.

So acid B is phthalic acid (ortho-benzenedicarboxylic acid).

Step 2: Identify compound A.

Since A gives a positive iodoform test, it must contain a $$-COCH_3$$ (methyl ketone) group. The iodoform test is positive for compounds containing the $$CH_3CO-$$ group.

The molecular formula of A is $$C_9H_{10}O$$. The degree of unsaturation is:

$$\text{DoU} = \frac{2(9) + 2 - 10}{2} = \frac{10}{2} = 5$$

A benzene ring accounts for 4 degrees of unsaturation (3 double bonds + 1 ring), and the $$C=O$$ of the ketone accounts for 1 more. Total = 5. So A is a substituted acetophenone (phenyl methyl ketone) with an additional methyl group on the ring.

Compound A = methyl acetophenone = $$CH_3-C_6H_4-COCH_3$$.

Step 3: Determine the position of the methyl group.

When A is oxidized with $$KMnO_4/KOH$$, both the $$-CH_3$$ group on the ring and the $$-COCH_3$$ group are oxidized to $$-COOH$$ groups.

The product B is phthalic acid, which has the two $$-COOH$$ groups in the ortho (1,2-) position.

This means the $$-CH_3$$ and $$-COCH_3$$ groups on A must be at ortho positions relative to each other.

Therefore, compound A is ortho-methyl acetophenone (2-methylacetophenone or 1-(2-methylphenyl)ethanone).

The answer is Option B: ortho-methyl acetophenone.