From Free Body Diagram of Block A

Applying static equillibrium equation in y direction,

$\sum F_y = 0 \\ \therefore N_A - 40\times 9.81 + T \sin{30 }= 0 \\ \therefore N_A + 0.5T = 392.4$

Also for x direction,

$\sum{F_x} = 0 \\ \therefore 0.3N_A- T \cos{30} = 0 \\ \therefore 0.3N_A - 0.866T = 0 $

Solving above two equations we get,

$N_A = 334.47N \text{ , }T=115.87N$

Also from free body diagram of Block B

Applying static equillibrium equation in y direction,

$\sum{F_y} = 0 \\ \therefore N_B-N_A-60\times 9.81 = 0 \\ \therefore N_B= 334.47+588.6 = 923.07 N $

and for x direction,

$\sum{F_x} = 0 \\ \therefore P- 0.3N_A-0.25N_B=0 \\ \therefore P = 0.3\times 334.07+0.25\times 923.07 = 331.11N$

Answer :

Force required to pull block B = 331.11N