To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.

The de-Broglie wavelength $$\lambda$$ is given by:

$$\lambda = \frac{h}{p}$$

where $$h$$ is Planck’s constant and $$p$$ is the momentum.

Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:

$$\lambda_{\text{electron}} = \lambda_{\text{proton}} \implies \frac{h}{p_{\text{e}}} = \frac{h}{p_{\text{p}}} \implies p_{\text{e}} = p_{\text{p}}$$

Next, the kinetic energy $$K$$ of a particle is related to its momentum $$p$$ and mass $$m$$ by the following formula:

$$K = \frac{p^2}{2m}$$

Given that the momentum $$p$$ is the same for both the proton and the electron, we can write the kinetic energies as:

$$K_{\text{e}} = \frac{p^2}{2m_{\text{e}}}$$

$$K_{\text{p}} = \frac{p^2}{2m_{\text{p}}}$$

The ratio of the kinetic energies is therefore:

$$\frac{K_{\text{e}}}{K_{\text{p}}} = \frac{\frac{p^2}{2m_{\text{e}}}}{\frac{p^2}{2m_{\text{p}}}} = \frac{m_{\text{p}}}{m_{\text{e}}}$$

Given that the mass of the proton $$m_{\text{p}}$$ is 1836 times the mass of the electron $$m_{\text{e}}$$, we have:

$$\frac{m_{\text{p}}}{m_{\text{e}}} = 1836$$

Thus, the ratio of the kinetic energies is:

$$\frac{K_{\text{e}}}{K_{\text{p}}} = 1836$$

Therefore, the correct answer is:

Option C

$$1: 1836$$