Explanation

(c)
Moles of $A,\ n_{A} = \frac{p_{A}V_{A}}{RT} = \frac{8\ \times \ 12}{RT} = \frac{96}{RT}$
Moles of $B,\ n_{B} = \ \frac{p_{B}V_{B}}{RT} = \frac{8\ \times \ 5}{RT} = \frac{40}{RT}$
Total pressure $\times$ total volume $= \left( n_{A} + n_{B} \right) \times RT$
$p \times (12 + 8) = \frac{1}{RT}(96 + 40)RT$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p = 6.8$
$\therefore Partial\ pressure\ of\ A = p \times mole\ fraction\ of\ A$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6.8 \times \left( \frac{96}{RT}/\frac{96 + 40}{RT} \right)\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4.8\ atm$
$\therefore Partial\ pressure\ of\ B = p \times mole\ fraction\ of\ B$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6.8\left( \frac{40}{RT}/\frac{96 + 40}{RT} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\ atm.$