Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the potential inside a parallel-plate capacitor is given by
$$V=Es$$
where $s$ is measured from the negative plate.
Hence,
$$V_{max}=Ed=500\;\rm V$$
And since the proton is fired from the middle point between the two plates,
$$V_i=250\;\rm V\tag 1$$
And since the energy is conserved, we need to find the initial speed that makes the proton reach the positive plate.
The energy is conserved, so
$$E_1=E_2$$
$$U_1+K_1=U_2+K_2$$
where $_1\rightarrow$ is at the middle point between the two plates and $_2\rightarrow$ is at the positive plate at which the full energy of the proton now is potential energy.
$$qV_i+K_1=qV_{max}+0$$
where $q=e$ which is the charge of the proton.
$$eV_i+\frac{1}{2}mv_1^2=eV_{max} $$
$$ v_1^2=2\left[\dfrac{eV_{max} -eV_i}{m}\right]$$
$$ v_1 =\sqrt{2e\left[\dfrac{ V_{max} - V_i}{m}\right]}$$
Plug from (1) and the rest of the known;
$$ v_1 =\sqrt{2(1.6\times 10^{-19})\left[\dfrac{500 -250}{1.67\times 10^{-27}}\right]}$$
$$v_1=\bf 218,870\;\rm m/s$$
which means that the speed of 200,000 m/s is not enough for the proton to reach the positive plate.
$$\color{blue}{\bf [b]}$$
The energy is conserved, so
$$E_1=E_3$$
$$U_1+K_1=U_3+K_3$$
where $_1\rightarrow$ is at the middle point between the two plates and $_3\rightarrow$ is at the negative.
$$eV_i+\frac{1}{2}mv_1^2=eV_{\rm min}+\frac{1}{2}mv_3^2$$
$$eV_i+\frac{1}{2}mv_1^2=0+\frac{1}{2}mv_3^2$$
Solving for $v_3$;
$$v_3=\sqrt{\dfrac{2eV_i}{m}+ v_1^2}$$
Plug the known;
$$v_3=\sqrt{\dfrac{2(1.6\times 10^{-19})(250)}{1.67\times 10^{-27}}+ (200,000)^2}$$
$$v_3=\color{red}{\bf 2.96\times 10^5}\;\rm m/s$$