Proton (P) and electron (e) will have same de-Broglie wavelength. Then the ratio of their kinetic energy is (assume, $m_p=1849m_e$ )

de-Broglie wavelength,  $\lambda =\frac{h}{p}$
Given:  $\frac{{\lambda }_{proton}}{{\lambda }_{electron}}=1$

$\frac{h/P_{proton}}{h/P_{electron}}=1\Rightarrow \frac{P_{proton}}{P_{electron}}=1 \Rightarrow \sqrt{\frac{2m_p{k}_p}{2m_e{k}_e}}=1$

$\Rightarrow \frac{k_p}{k_e}=\frac{m_e}{m_p}=\frac{m_e}{1849m_e} =1:1849$