Since $P=x_AP_A^0+x_BP_B^0$
$\Rightarrow \big(\large\frac{1}{1+3}\big)$$P_A^0+\big(\large\frac{3}{1+3}\big)$$P_B^{\large\circ}$=550mm Hg
$\Rightarrow \big(\large\frac{1}{1+4}\big)$$P_A^0+\big(\large\frac{4}{1+4}\big)$$P_B^{\large\circ}$=560mm Hg
That is,
$0.25P_A^0+0.75P_B^0$=550mm Hg
$0.20P_A^0+0.8P_B^0$=560mm Hg
Solving for $P_A^0$ and $P_B^0$ we get,
$P_A^0=400$ mm Hg
$P_B^0=600$ mm Hg