Dihaloalkane reaction with sodamide/ammonia In the following sequence of reactions:
$$\ce{CH3-CH2-CH2-I -&gt;[\text{Alc.}~KOH] \textbf{A} -&gt;[Br2] \textbf{B} -&gt;[NaNH2/NH3] \textbf{C}}$$
the product C is:

alkene
alkanol
alkyne
alkylamine


I haven't really came across a reagent $\ce{NaNH2}/\ce{NH3}$ in my whole session. Well, I have got B which is $\ce{CH3-CH(Br)-CH2-Br}.$
Then, how do I proceed next? How will my substrate  react with that reagent and what product will I achieve?
You were correct up to this point, the strong base alcoholic KOH sets off elimination instead of substitution so you will not get an alcohol.
Accordingly your products A (presumably propene) and B are correct.
The amide/ammonia is an even stronger base than alcoholic KOH and will also produce elimination.  With a vicinal dihalide like your B, there is an added twist.  Read here to find out what happens.