To determine the correct relation between the kinetic energies of a proton ($$\mathrm{K}_{\mathrm{p}}$$) and an electron ($$\mathrm{K}_{\mathrm{e}}$$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:

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

where $$\lambda$$ is the de Broglie wavelength, $$h$$ is Planck's constant, and $$p$$ is the momentum of the particle.

The momentum $$p$$ of a particle is given by:

$$p = \sqrt{2mK}$$

where $$m$$ is the mass of the particle and $$K$$ is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:

$$\lambda_{\mathrm{p}} = \lambda_{\mathrm{e}}$$

This implies the momenta should be the same:

$$p_{\mathrm{p}} = p_{\mathrm{e}}$$

Thus, we can write:

$$\sqrt{2 m_{\mathrm{p}} K_{\mathrm{p}}} = \sqrt{2 m_{\mathrm{e}} K_{\mathrm{e}}}$$

Squaring both sides to eliminate the square roots:

$$2 m_{\mathrm{p}} K_{\mathrm{p}} = 2 m_{\mathrm{e}} K_{\mathrm{e}}$$

$$m_{\mathrm{p}} K_{\mathrm{p}} = m_{\mathrm{e}} K_{\mathrm{e}}$$

Rearranging to solve for $$K_{\mathrm{p}}$$ in terms of $$K_{\mathrm{e}}$$:

$$K_{\mathrm{p}} = \frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} K_{\mathrm{e}}$$

Since the mass of a proton $$m_{\mathrm{p}}$$ is much greater than the mass of an electron $$m_{\mathrm{e}}$$:

$$m_{\mathrm{p}} \gg m_{\mathrm{e}}$$

This means:

$$\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} \ll 1$$

Therefore, it implies:

$$K_{\mathrm{p}} < K_{\mathrm{e}}$$

So, the correct option is:

Option C: $$\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$$