Let $\phi\colon G\to K$ be an epimorphism and let $J\lhd K$. Show that there is a normal subgroup $H\lhd G$ so that $\frac{G}{H}\cong \frac{K}{J}$.
Since $\phi\colon G\to K$ is an epimorphism, by a proposition that I have already seen in class, there is a one-to-one correspondence between the subgroups of $K$ and the subgroups of $G$ that contain $\ker(\phi)$, given by $H\mapsto \phi(H)$. We have $H\lhd G$ if and only if $\phi(H)\lhd K$.
Take this $H$ to be a normal subgroup of $G$ that contains the $\ker(\phi)$, i.e. $\ker(\phi)\lhd H\lhd G$. Since $H\lhd G$, we know that $\phi(H)\lhd K$, by the theorem.
Not finished yet
Consider the canonical projection $\pi\colon K\to K/J$. Then you know from the homomorphism theorems that $$ K/J \cong G/\ker(\pi\circ\phi) $$ because $\pi\circ\phi$ is surjective.