De Broglie wavelength $\lambda = h/p = h/\sqrt{2mK}$. So, kinetic energy $K \propto 1/\lambda^2$. If the wavelength is halved, the new kinetic energy $K'$ will be $1/(1/2)^2 = 4$ times the initial energy $K$. The energy added is $K' - K = 4K - K = 3K$, which is thrice the initial energy.