I can prove in Coh(P^1) there isn't enough projective objects, either, by using Serre's duality theorem and vanishing theorem. But I don't know this is the case for Qcoh(P^1).

I can prove that in $\operatorname{Coh}(\boldsymbol{P}^1)$ there aren't enough projective objects by using Serre's duality theorem and some vanishing theorems. But I don't know if this is the case for $\operatorname{Qcoh}(\boldsymbol{P}^1)$.