From the theory of special relativity, \begin{align} E = \sqrt{p^2c^2 + m^2c^4}. \end{align} Since the photon is a massless particle, and \begin{align} E = pc. \end{align}

Then, \begin{align} \hbar &= \frac{h}{2\pi}\\ \nu &= \frac{h}{2\pi}\\ \omega &= 2\pi\nu\\ k &=\frac{1}{\lamda}\\ p &={\hbar}k\\ E &=h\nu=2\pi\hbar\nu=\hbar\omega\\ E &=pc={\hbar}kc=\frac{{\hbar}c}{\lamda}\\ \end{align}