Explanation

(c) Displacement of the man with respect to trolley in $4\ sec$
$x_{mT} = 4m \Rightarrow x_{mT} = x_{m} + x_{T} \Rightarrow x_{T} = 4 - x_{m}$
Position of centre of mass remain constant
$\Rightarrow \left( 4 - x_{m} \right)320 = x_{m} \times 80 \Rightarrow x_{m} = \frac{16}{5} = 3.2\ m$
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Alternatively: If the man starts walking on the trolley in the forward direction then whole system will move in backward direction with same momentum.
Momentum of man in forward direction = Momentum of system (man + trolley) in backward direction
$\Rightarrow 80 \times 1 = (80 + 320) \times v \Rightarrow v = 0.2\ m/s$
So the velocity of man w.r.t. ground $1.0 - 0.2 = 0.8\ m/s$
$\therefore$ Displacement of man w.r.t. ground $= 0.8 \times 4 = 3.2\ m$