The correct match between column-I and column-II is:
Column-I (drug)              Column-II (test)
(A) Chloroxylenol           (P) Carbylamine test
(B) Norethindrone         (Q) Sodium hydrogen carbonate test
(C) Sulphapyridine        (R) Ferric chloride test
(D) Penicillin                  (S) Baeyer's test

First, we recall the characteristic qualitative tests that are commonly used in organic chemistry:

$$\text{Ferric chloride test}$$ is given by compounds containing the phenolic $$-OH$$ group, producing a coloured complex.

$$\text{Carbylamine test}$$ is positive for aliphatic or aromatic primary amines. When the compound is warmed with chloroform and alcoholic $$KOH$$, the foul-smelling isocyanide (carbylamine) is formed.

$$\text{Sodium hydrogen carbonate (NaHCO}_3\text{) test}$$ detects carboxylic acids. The acid reacts with $$NaHCO_3$$ to liberate $$CO_2$$ gas, visible as brisk effervescence.

$$\text{Baeyer’s test}$$ employs dilute alkaline $$KMnO_4$$ solution. Compounds containing unsaturation (double or triple bonds) decolourise the purple $$KMnO_4$$ solution while producing a brown precipitate of $$MnO_2$$.

Now we examine each drug in Column-I one by one and match it with the appropriate test from Column-II.

We have Chloroxylenol. Its structure possesses a phenolic $$-OH$$ group attached to an aromatic ring. A phenolic group responds positively to the ferric chloride test, forming a violet to blue-green complex. So, for Chloroxylenol:

$$\text{Chloroxylenol} \longrightarrow \text{Ferric chloride test}$$

Hence, $$A \rightarrow R$$.

Next, consider Norethindrone. This steroidal contraceptive molecule contains an ethynyl $$(-C \equiv CH)$$ group, i.e., a carbon-carbon triple bond. According to Baeyer’s test, any compound with such unsaturation reacts with dilute alkaline $$KMnO_4$$ and discharges its purple colour. Therefore, for Norethindrone:

$$\text{Norethindrone} \longrightarrow \text{Baeyer’s test}$$

Thus, $$B \rightarrow S$$.

Now, we take Sulphapyridine. The ring nitrogen is bonded to a primary $$-NH_2$$ group on an aromatic ring. A primary amine gives the carbylamine reaction:

$$\text{ArNH}_2 + CHCl_3 + 3\,KOH \;\longrightarrow\; \text{ArNC} + 3\,KCl + 3\,H_2O$$

The unpleasant odour of isocyanide confirms the presence of a primary amine. Hence, Sulphapyridine matches the carbylamine test:

$$\text{Sulphapyridine} \longrightarrow \text{Carbylamine test}$$

Therefore, $$C \rightarrow P$$.

Finally, we analyse Penicillin. Every penicillin molecule carries a free carboxylic acid $$-COOH$$ group. When this acidic hydrogen reacts with sodium hydrogen carbonate, carbon dioxide gas is evolved:

$$RCOOH + NaHCO_{3} \rightarrow RCOONa + CO_{2} \uparrow + H_{2}O$$

The effervescence confirms a positive NaHCO3 test. So, Penicillin corresponds to the sodium hydrogen carbonate test:

$$\text{Penicillin} \longrightarrow \text{Sodium hydrogen carbonate test}$$

Consequently, $$D \rightarrow Q$$.

Collecting all the results, we obtain the complete matching:

$$A \rightarrow R,\; B \rightarrow S,\; C \rightarrow P,\; D \rightarrow Q$$

This sequence is exactly the same as Option D.

Hence, the correct answer is Option D.