To see that condition d implies condition a, assume that condition d holds. Then $b^{-1}a \in H$ also, because $H$ is closed under inverses. Now consider an arbitrary element $ah$ of $aH$, where $h \in H$. Then $ah = (b(b^{-1}a))h = b((b^{-1}a)h) \in bH$. Therefore $aH \subseteq bH$. For the converse, consider an arbitrary element $bh$ of $bH$, where $h \in H$. we have $bh = (a(a^{-1}b))h = a((a^{-1}b)h) \in aH$. Therefore $bH \subseteq aH$.
(I don't remember why I put the hypothesis of normality in the question.You could use it to get equivalent statements like $aH = Hb$, or $aHb^{-1} = H$, but I don't think these are especially interesting. Perhaps I meant to ask about conditions equivalent to the normality of $H$; you should know several of those also.)