One mole each of He and A(g) are taken in a 10 L closed flask and...

The given reaction is $A(g) \rightleftharpoons B(g)$.

Let the initial moles of $A$ be $1$ and at equilibrium, let $x$ moles of $A$ dissociate.

Moles at equilibrium:
$n_A = 1 - x$
$n_B = x$

Since $\Delta n_g = 0$, the equilibrium constant $K_c$ can be written in terms of moles:
$K_c = \dfrac{n_B}{n_A} = \dfrac{x}{1 - x} = 4.0$

$x = 4 - 4x \Rightarrow 5x = 4 \Rightarrow x = 0.8$

At equilibrium, the moles of the gases are:
$n_{He} = 1$
$n_B = 0.8$

Using the ideal gas equation $P = \dfrac{nRT}{V}$, the partial pressures are calculated as follows:

Partial pressure of He:
$P_{He} = \dfrac{1 \times 0.082 \times 400}{10} = 3.28$ atm

Partial pressure of $B(g)$:
$P_B = \dfrac{0.8 \times 0.082 \times 400}{10} = 2.624$ atm

Answer: $3.28, 2.624$