In question, it is asked for the conditional probability.
$\mathbb{P}(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ where $\lambda \in (0,\infty)$ and $k \in N_0$ (Natural numbers starts from 0)
For $\lambda=1,$ $\mathbb{P}(X=k) = \frac{1}{ek!}$
Now, $\mathbb{P}(X \geq 2 \ | \ X \leq 4 ) = \frac{\mathbb{P} (X \geq 2 \cap X \leq 4)}{\mathbb{P} (x \leq 4)}$
$= \frac{\mathbb{P}(X=2)+ \mathbb{P}(X=3) + \mathbb{P}(X=4)}{\mathbb{P}(X=0)+ \mathbb{P}(X=1)+ \mathbb{P}(X=2)+\mathbb{P}(X=3) + \mathbb{P}(X=4)}$