1. Identify compound $P$ from cumene oxidation

Cumene undergoes air oxidation followed by acidic hydrolysis by the cumene process:

Since the next step is with $CHCl_3/NaOH$ (Reimer–Tiemann reaction), the reactive component must be phenol.

So,

$P = \text{phenol}$

2. Reaction of phenol with $CHCl_3/NaOH$

Phenol with $CHCl_3/NaOH$ undergoes Reimer–Tiemann reaction to give mainly o-hydroxybenzaldehyde and a small amount of p-hydroxybenzaldehyde.

Thus,

$Q = o\text{-hydroxybenzaldehyde} \quad (\text{major})$ $R = p\text{-hydroxybenzaldehyde} \quad (\text{minor})$

3. Formation of $S$ from $Q$ with $PhCH_2Br/NaOH$

$Q$ is $o$-hydroxybenzaldehyde. In presence of $NaOH$, the phenolic $-OH$ forms phenoxide and reacts with benzyl bromide via Williamson ether synthesis:

So $S$ is o-benzyloxybenzaldehyde.

4. Check each option

Option A: $R$ is steam volatile.

$R$ is $p$-hydroxybenzaldehyde. Steam volatility is shown prominently by o-hydroxybenzaldehyde due to intramolecular H-bonding and lower association. The para isomer forms intermolecular H-bonding and is not steam volatile.

So, A is false.

Option B: $Q$ gives dark violet coloration with 1% aqueous $FeCl_3$ solution.

$Q$ is $o$-hydroxybenzaldehyde, which still contains a phenolic $-OH$ group. Phenols give colored complexes with neutral $FeCl_3$.

Hence, B is true.

Option C: $S$ gives yellow precipitate with 2,4-dinitrophenylhydrazine.

$S$ is $o$-benzyloxybenzaldehyde. It contains an aldehyde group.

Aldehydes and ketones give yellow/orange precipitate with 2,4-DNP reagent.

Hence, C is true.

Option D: $S$ gives dark violet coloration with 1% aqueous $FeCl_3$ solution.

In $S$, the phenolic $-OH$ has been converted into a benzyl ether ($-OCH_2Ph$). Therefore, there is no free phenolic group left.

So it will not give the $FeCl_3$ coloration test.

Hence, D is false.

5. Final answer

Correct statements are:

$\boxed{B,\ C}$

6. Comparison with stored correct answer

Stored correct answer: $B, C$

My derived answer matches it exactly.