Answer
a) $\text{Into a region of higher potential.}$
b) $3340\;\rm V$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the proton slows down as it travels and we know that the proton is a positively charged particle. This means that the electron is moving toward a higher potential region.
$$\color{blue}{\bf [b]}$$
We know that the potential difference is given by
$$\Delta V=\dfrac{\Delta U}{q_p}\tag 1$$
And since the energy is conserved,
$$E_i=E_f$$
$$K_i+U_i=K_f+U_f$$
The final speed of the electron is zero, so $K_f=0$
$$K_i+U_i=0+U_f$$
Hence,
$$\Delta U=K_i=\frac{1}{2}m_pv_p^2$$
Plug into (1),
$$\Delta V=\dfrac{\frac{1}{2}m_pv_p^2}{q_p} $$
Plug the known;
$$\Delta V=\dfrac{\frac{1}{2}(1.67\times 10^{-27})(800,000)^2}{(1.6\times 10^{-19})} $$
$$\Delta V=\color{red}{\bf 3340}\;\rm V$$
