Answer
The speed of the electron relative to the lab is $\frac{1}{5}~c$
Work Step by Step
Let $v_{PL}$ be the velocity of the proton relative to lab. Then $v_{PL} = \frac{4}{5}~c$
Let $v_{EP}$ be the velocity of electron relative to the proton. Then $v_{EP} = -\frac{5}{7}~c$
We can find $v_{EL}$:
$v_{EL} = \frac{V_{EP}~+~v_{PL}}{1+\frac{(v_{EP})~(v_{PL})}{c^2}}$
$v_{EL} = \frac{(-\frac{5}{7}~c)~+~(\frac{4}{5}~c)}{1+\frac{(-\frac{5}{7}~c)~(\frac{4}{5}~c)}{c^2}}$
$v_{EL} = \frac{\frac{3}{35}~c}{\frac{3}{7}}$
$v_{EL} = \frac{1}{5}~c$
The velocity of the electron relative to the lab is $\frac{1}{5}~c$
Therefore, the speed of the electron relative to the lab is $\frac{1}{5}~c$
(to the right).
