I am not sure what the question is really about but I will give a counter-example as I understood. Consider $S_3$ and set $G=\{1,(12)\},H=\{1,(123),(132)\}$ and $K=\{1,(13)\}$. Then $HK=S_3$ so $G\cap HK=G$ but $G\cap H=G\cap K=1$ so we get the trivial subgroup on the right hand side.
In general, the right hand side is always a subgroup of the left hand side but the equality holds in some special cases. You can check Dedekind's Lemma.
I am not sure what the question is really about but I will give a counter-example as I understood. Consider $S_3$ and set $G=\{1,(12)\},H=\{1,(123),(132)\}$ and $K=\{1,(13)\}$. Then $HK=S_3$ so $G\cap HK=G$ but $G\cap H=G\cap K=1$ so we get the trivial subgroup on the right hand side.
In general, the right hand side is always a subgroup of the left hand side but the equality holds in some special cases. You can check Dedekind's Lemma.