You're forgetting that an ammonium cation is formed which is mildly acidic and will easily rid one of it's protons, reforming it's lone pair to continue attacking electrophiles. In each case, the e-donating contribution of alkyl substituents increase the electron density of the lone pair (it's nucleophilicity); however, for the same reason, it becomes more difficult to remove a proton (less acidic due to increasing instability/reactivity). It'll continue attacking electrophiles until no more protons remain, which is why D is the correct answer. Technically, it could even go one further step and create an alkyl ammonium ion.
but why does it trade the hydrogen for an ethyl group? is it because carbon is more electrophilic/electronegative?
It doesn't trade it for a hydrogen. The NH3 has a free lone pair to attack the electrophile and displace the leaving group. Once the alkyl ammonium is formed, it'll rid one of it's protons via deprotonation because it's mildly acidic. This "free's up" the lone pair again to repeat the process.
Compared to a neutral amine, it is not. Remember alkyl substituents donate some electron density inductively, which in turn makes it's free lone pair more negative and therefore, more reactive. The key part of this question is what would happen if excess electrophile was added. Sure, after one reaction it'd stop but provided there's so much electrophile floating around, it'll keep reacting as much as possible.
makes sense thanks!
