vapour pressure of $B=P_B^o$

In first solution,

Mole fraction of $A\,({x_A})\, = \,\frac{1}{{1 + 2}}\, = \,\frac{1}{3}$

Mole fraction of $B\,({x_B})\, = \,\frac{2}{{1 + 2}}\, = \,\frac{2}{3}$

According to Raoult's law, 

Total vapour pressure

$ = \,250\, = \,P_A^o{x_A}\, + \,P_B^o{x_B}$

$\,250\, = \,\frac{1}{3}P_A^o + \,\frac{2}{3}P_B^o$   ....... $(i)$

In second solution

Mole fraction of $A\,({x_A})\, = \,\frac{2}{{2\, + \,2}}\, = \,\frac{2}{4}\, = \frac{1}{2}$

Mole fraction of $B({x_B})\,\, = \,\frac{2}{4}\, = \frac{1}{2}$

$\therefore $ Total vapour pressure

$ = \,300\, = \,P_A^o{x_A}\, + \,P_B^o{x_B}$

$300\, = \,\frac{1}{2}P_A^o\, + \,\frac{1}{2}P_B^o$   ..... $(ii)$

Multiplying equation $(i)$ by $\frac {1}{2}$ and equation $(ii)$ by $\frac {1}{3}$

$\frac{1}{6}P_A^o\, + \,\frac{2}{6}P_B^o\, = \,125$

$\frac{1}{6}P_A^o\, + \,\frac{1}{6}P_B^o\, = \,100$

$\frac{1}{6}P_B^o\, = \,25$

$P_B^o\, = \,25\, \times \,6\, = \,150\,mm\,Hg$

On substituting value of $P_B^o$ in equation $(ii)$ we get

$300\, = \,P_A^o\, \times \,\frac{1}{2}\, + \,150\, \times \frac{1}{2}$

$\,P_A^o\, = \,450\,mm\,Hg$