Dihaloalkane reaction with sodamide/ammonia
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.