Which formula for the de Broglie wavelength of an electron is correct?

Often, when dealing with high-energy (relativistic) particles the rest mass of the particle can be neglected when performing calculations. Use your expression for $p$ from relativistic considerations, plug in the numbers and see the negligible change when you include and neglect to include the mass of the electron.

A good tip for when you enter into higher level physics/astrophysics: approximations are made all the time, where higher order terms will ve neglected.

As the energy of the electrons in that case is much greater than their mass, you can consider the approximation $E \sim pc$. So the formulas are equivalent.

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 $E\gg m_0c^2$, thus this yields $p=\frac{E}{c}$.