(i) When white phosphorus is heated with concentrated $\mathrm{NaOH}$ in inert atmosphere of $\mathrm{CO}_2$, it gives $\mathrm{PH}_3(g)$ and $\mathrm{NaH}_2 \mathrm{PO}_2$ as follows :
$$
\begin{aligned}
& \mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \stackrel{\mathrm{CO}_2}{\longrightarrow} \underset{(A)}{\mathrm{PH}_3(g)} \\
& +\quad 3 \mathrm{NaH}_2 \mathrm{PO}_4 \\
& \text { (Sodium hypophosphite) } \\
&
\end{aligned}
$$
(B)
ii) When $(A)$, i.e. $\mathrm{PH}_3(g)$ is bubbled into aqueous $\mathrm{CuSO}_4$ solution, copper phosphide $\left(\mathrm{Cu}_3 \mathrm{P}_2\right)$ and $\mathrm{H}_2 \mathrm{SO}_4(\mathrm{C})$ is formed. The reaction occurs as follows :
$$
2 \mathrm{PH}_3+3 \mathrm{CuSO}_4 \longrightarrow \underset{\begin{array}{c}
(\text { Copper } \\
\text { phosphide) }
\end{array}}{\mathrm{Cu}_3 \mathrm{P}_2}+3 \mathrm{H}_2 \mathrm{SO}_4
$$

Hence, $(B)=\mathrm{NaH}_2 \mathrm{PO}_2,(C)=\mathrm{H}_2 \mathrm{SO}_4$
Hence, option (b) is the correct answer.