Answer
a) $4\times 10^5\;\rm Pa$
b) $\rm Irreversible.$
Work Step by Step
First of all, we need to find the number of moles of each gas, and since both containers are sealing ideal gases, we need to use the ideal gas law.
$$P_{1A}V_{A}=n_{1A}RT_{A}$$
Hence,
$$n_{1A}=\dfrac{P_{1A}V_{A}}{ RT_{A}}\tag 1$$
And by the same approach,
$$n_{1B}=\dfrac{P_{1B}V_{B}}{ RT_{B}} $$
We know that $B_B=4V_A$, so
$$n_{1B}=\dfrac{4P_{1B}V_{A}}{ RT_{B}} $$
We also know, from the given data, that $P_{1B}=5P_{1A}$
$$n_{1B}=\dfrac{20P_{1A}V_{A}}{ RT_{1B}} $$
Also, $T_{A}/T_{B}=3/4$, so $T_{B}=\frac{4}{3}T_{A}$,
$$n_{1B}=\dfrac{20P_{1A}V_{A}}{\frac{4}{3} RT_{A}} =15\left(\dfrac{P_{1A}V_{A}}{ RT_{A}}\right)$$
Plugging from (1);
$$n_{1B}=15\;n_{1A}\tag 2$$
a) Now when the valve has opened the pressure of both gases, after a while, will be the same in both containers.
Thus, the final pressure of each gas is equal but the number of moles of the gas in each container differs since some will move from one container to the other to equalize the pressure.
$$P_{2A}=P_{2B}$$
Thus,
$$\dfrac{ n_{2A}RT_{A}}{ V_{1A}}=\dfrac{ n_{2B}RT_{B}}{ V_{1B}}$$
Noting that the temperature of each container remains constant by heaters as the author told us.
$$\dfrac{ n_{2A}\color{red}{\bf\not} R\color{red}{\bf\not} T_{A}}{ \color{red}{\bf\not} V_{ A}}=\dfrac{ n_{2B}\color{red}{\bf\not} R(\frac{4}{3}\color{red}{\bf\not} T_A)}{ 4\color{red}{\bf\not} V_{ A}}$$
Hence,
$$n_{2B}=3\;n_{2A}\tag 3$$
Noting that
$$n_{1A}+n_{1B}=n_{2A}+n_{2B}$$
Plugging from (2) and (3);
$$n_{1A}+15\;n_{1A}=n_{2A}+3\;n_{2A}$$
Thus,
$$n_{2A}=4n_{1A}\tag 4$$
__________________________________________________________
a) Now we can find the new pressure of any gas which is the pressure of both of them,
$$P_2=\dfrac{n_{2A}RT_A}{V_A}$$
Solving (1) for $V_A$ and plug it here;
$$P_2=n_{2A}RT_A\left[\dfrac{1}{V_A}\right]=n_{2A}\color{red}{\bf\not} R\color{red}{\bf\not} T_A\left[\dfrac{P_{1A}}{ n_{1A}\color{red}{\bf\not} R\color{red}{\bf\not} T_{A}}\right] $$
$$P_2=\dfrac{n_{2A}}{n_{1A}}P_1$$
Plugging from (4);
$$P_2=\dfrac{4\color{red}{\bf\not} n_{1A}}{\color{red}{\bf\not} n_{1A}}P_1=4P_1$$
Therefore,
$$P_2=\color{red}{\bf 4\times 10^5}\;\rm Pa$$
__________________________________________________________
b) This process of opening the valve is irreversible. We can't return everything to the initial state before opening the valve. The gas spreads out between the two containers and is mixed together until it reached an equilibrium point. So it is irreversible.
