In the following sequence of reactions, the final product D is:

The reaction sequence can be understood by analyzing each step individually.

In the first step, propyne,

$$\mathrm{CH_3-C\equiv C-H},$$

is treated with sodium amide,

$$\mathrm{NaNH_2}.$$

Since propyne is a terminal alkyne, the hydrogen attached to the triple-bonded carbon is acidic. The strong base $$\mathrm{NH_2^-}$$ abstracts this proton to form the corresponding acetylide ion.

Thus, intermediate A is sodium propynide,

$$\boxed{\mathrm{CH_3-C\equiv C^-Na^+.}}$$

In the second step, the acetylide ion reacts with 4-bromobutan-2-ol,

$$\mathrm{Br-CH_2-CH_2-CH(OH)-CH_3}.$$

The acetylide ion acts as a nucleophile and attacks the primary carbon bearing the bromine atom through an $$S_N2$$ mechanism, displacing $$\mathrm{Br^-}$$ and forming a new carbon-carbon bond.

The resulting intermediate B is

$$\boxed{\mathrm{CH_3-C\equiv C-CH_2-CH_2-CH(OH)-CH_3}.}$$

In the third step, intermediate B is treated with

$$\mathrm{H_2/Pd-C}.$$

Catalytic hydrogenation completely reduces the carbon-carbon triple bond to a single bond without affecting the alcohol group.

Hence, intermediate C becomes

$$\boxed{\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-CH(OH)-CH_3},}$$

which is heptan-2-ol.

In the final step, intermediate C is oxidized with

$$\mathrm{CrO_3}.$$

Chromium trioxide converts a secondary alcohol into the corresponding ketone.

Therefore, the hydroxyl-bearing carbon is oxidized to a carbonyl group, giving

$$\boxed{\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-C(=O)-CH_3}.}$$

This compound is heptan-2-one.

Hence, the final product D is

$$\boxed{\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-C(=O)-CH_3}.}$$

Therefore, the correct answer is

$$\boxed{\text{Option (A).}}$$