 Solution

$\boxed{\text{No.}}$ A disjunctive Datalog knowledge base can have several incomparable minimal Herbrand models without a least Herbrand model. Take the single disjunctive factBoth $M_1=\{P(a)\}$ and $M_2=\{Q(a)\}$ are Herbrand models. Any least Herbrand model would have to be contained in both, hence contained in $M_1\cap M_2=\varnothing$. But the empty Herbrand interpretation falsifies the disjunctive fact. This is a contradiction. Taking the intersection of all models therefore fails because that intersection need not itself be a model.