Which one of the following reaction sequences will give an azo dye?

An azo dye contains the chromophore $$\,\text{-N}=\text{N-}\,$$ that is created by coupling an aromatic diazonium ion with an electron-rich aromatic ring (phenol, aniline, $$\beta$$-naphthol, etc.).
Hence two essential stages are required:

1. Diazo­tisation: convert a primary aromatic amine to a diazonium salt with $$NaNO_2 + HCl$$ at $$0{-}5^{\circ}C$$.
2. Coupling: react that diazonium salt, in weakly alkaline (or mildly acidic) medium, with a strongly activated aromatic ring to give $$Ar{-}N=N{-}Ar'$$, which is coloured.

Option A:
$$C_6H_5NH_2 \xrightarrow[\;0-5^{\circ}C\;]{NaNO_2/HCl} C_6H_5N_2^+Cl^-$$ (benzenediazonium chloride)
$$C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow[\;278{-}283\;K\;]{NaOH} C_6H_5N=N{-}C_6H_4OH\;(\text{p-hydroxy­azobenzene,\;Orange I})$$

The product contains the $$\text{-N}=\text{N-}$$ linkage and is an intensely coloured azo dye. Therefore Option A satisfies both necessary steps and definitely yields an azo dye.

Why the other options fail (briefly):
• They either never generate an aromatic diazonium salt, or
• the intermediate produced is not allowed to couple with an activated ring, or
• further reactions (Sandmeyer, reduction, hydrolysis etc.) destroy the diazonium ion before coupling can occur.
Thus no $$\text{-N}=\text{N-}$$ chromophore, hence no azo dye.

Option A which is: diazotisation of aniline followed by alkaline coupling with phenol, is the correct choice.