Depends on the logical calculus you are using.

In words, the derivation is the following. Suppose (Ex) (Kx & ~Hx) is true. Take that x0 that makes it true; then K x0 & ~H x0, which gives K x0 and ~H x0 separately.

Then substitute x0 into (Ax) (Kx > Hx) to get K x0 -> H x0. By Modus Ponens and previously derived K x0 we get H x0. But we also have ~H x0, a contradiction. Therefore, (Ex) (Kx & ~Hx) is false, i.e., ~(Ex) (Kx & ~Hx).