If $$\lambda$$ and $$K$$ are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be

For a particle of constant mass $$m$$:

$$\lambda = \frac{h}{\sqrt{2mK}} \implies \lambda^2 = \frac{h^2}{2mK}$$

Isolating the variables on the axes ($$y = \frac{1}{K}$$ and $$x = \lambda$$):

$$\frac{1}{K} = \left(\frac{2m}{h^2}\right)\lambda^2$$

Comparing with standard parabola $$y = c x^2$$ where $$c > 0$$:

$$\frac{1}{K} \propto \lambda^2$$

This represents a parabolic curve opening upwards with its vertex at the origin $$(0,0)$$.