Step 1: Neutralise the acid first.
Phthalic acid has two $-COOH$ groups on adjacent ring positions. Ammonia, being basic, simply neutralises both of them on mixing, giving the diammonium salt, ammonium phthalate, $C_6H_4(COONH_4)_2$.
Step 2: Drive off water on mild heating.
Gentle heating pulls water molecules out of the ammonium salt, converting each $-COONH_4$ group into an amide group, $-CONH_2$. This gives phthalamide, $C_6H_4(CONH_2)_2$.
Step 3: Push the reaction further with strong heat.
Since the two amide groups sit right next to each other on the ring, stronger heating lets one $-CONH_2$ attack the other, expelling a further molecule of ammonia and closing a five-membered ring in the process.
\[ C_6H_4(COOH)_2 \xrightarrow{NH_3} C_6H_4(COONH_4)_2 \xrightarrow{\Delta} C_6H_4(CONH_2)_2 \xrightarrow{\text{strong }\Delta,\ -NH_3} \text{Phthalimide} \]
\[ \boxed{\text{Phthalimide}} \]