A person ( 40 kg ) manages to remain at rest between two vertical...

Let the normal force exerted by wall $A$ on the person be $N_A$ and by wall $B$ be $N_B$.

For horizontal equilibrium, the net horizontal force on the person must be zero. Therefore, the normal forces exerted by the two walls must be equal:
$N_A = N_B = N$

For vertical equilibrium, the total upward frictional force must balance the downward weight of the person. The weight of the person is $W = mg = 40 \times 10 = 400 \text{ N}$.

Since limiting friction acts at all contact points, the frictional forces at walls $A$ and $B$ are:
$f_A = \mu N_A = \mu N$
$f_B = \mu N_B = \mu N$

Equating the total upward frictional force to the weight:
$f_A + f_B = mg$
$\mu N + \mu N = mg$
$2\mu N = mg$

Substituting the given values ($\mu = 0.8$, $mg = 400 \text{ N}$):
$2 \times 0.8 \times N = 400$
$1.6 N = 400$
$N = \dfrac{400}{1.6} = 250 \text{ N}$

Thus, the normal force exerted by either wall on the person is $250 \text{ N}$.