The 4-momentum vector is given by ${\bf p}=(\frac{E}{c},p^{1},p^{2},p^{3})$. Now taking the scalar product with itself we have,
\begin{equation}
{\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4
\end{equation}
Now for extremely relativistic case , we can use the condition that $E>>m_0c^2$, thus this yields $p=\frac{E}{c}$.