The set $(Ha)(Hb)$ consists of all elements of the form $h_1ah_2b$, for $h_1,h_2\in H$.
Suppose $H$ is normal in $G$. We want to show that $(Ha)(Hb)=Hab$.
_Proof that $(Ha)(Hb)\subseteq Hab$_. Let $h_1,h_2\in H$. Then $h_1ah_2b=h_1ah_2a^{-1}ab$. If $h=h_1ah_2b=h_1ah_2a^{-1}=h_1(ah_2a^{-1})$ we see that, by normality, $h\in H$. Therefore $h_1ah_2b=hab\in Hab$.
_Proof that $Hab\subseteq (Ha)(Hb)$._ Let $h\in H$; then $hab=1a(a^{-1}ha)b$. Set $h_1=1\in H$ and $h_2=a^{-1}ha\in H$; then $hab=h_1ah_2b\in(Ha)(Hb)$
Suppose $(Ha)(Hb)=Hab$, for all $a,b\in G$. We want to show that $H$ is normal in $G$.
Let $g\in G$ and $h\in H$. Then $ghg^{-1}=1ghg^{-1}\in (Hg)(Hg^{-1})$. Since $(Hg)(Hg^{-1})=Hgg^{-1}$ we know there exists $h_0\in H$ such that $$ ghg^{-1}=h_0gg^{-1} $$ Therefore $ghg^{-1}=h_0\in H$.