I found a similar question already answered here: Find CDF of random variable which depends on other variable Our goal is to find $f_Y$ (the PDF) which is the derivative of $F_Y$ (the CDF).
$X$ is a random variable distributed as $N(0,1)$, thus the PDF is:
$f_X(x) = \frac{e^{-x^2/2}}{\sqrt{2\pi}}$
While the CDF is: $F_X(x) = \frac{1}{\sqrt{2\pi}}\int_{- \infty}^{x}e^{-t^2/2}\,dt$
The support of $Y$ is $[-1, 1]$. Clearly:
$P(Y = -1) = P(X \leq -2)$
$P(Y = 0) = P(-1 \leq X \leq 1)$
$P(Y = 1) = P(X \geq 2)$
Finally:
$P(-1 \lt Y \lt 0) = P(-2 \lt X \lt -1)$
$P(0 \lt Y \lt 1) = P(1 \lt X \lt 2)$
Though I have managed to break apart the CDF of $Y$, I'm unsure how to simplify these further. The CDF of $X$ is very complicated unlike in the question I've linked above.
The question is borrowed from GATE EC 2023.
It is totally unnecessary to perform any detailed computations.
Assuming $\delta$ represents the Dirac delta function centered at $0$, we can immediately see from the graph that $Y$ is a mixed discrete-continuous random variable with discrete probability masses at $Y \in \{-1, 0, 1\}$. Although the figure does not actually show whether the event $|Y| > 1$ has positive probability, we can infer from the location of these masses that the probability density of $Y$ must have delta functions at these locations. Choices (A) and (D) are missing the one at $y = 0$; choices (C) and (D) are missing the ones at $y = \pm 1$. Only choice (B) contains all three, so this must be the best answer.
However, as I have pointed out, what we can see of the graph of $h$ does not exclude the possibility that $h(X)$ may take on values outside the interval $[-1,1]$, so for instance, that there might be an interval for which $h(X) = 2$. It might be vaguely implied that for all $X \ge 2$ that $h(X) = 1$, and that for $X \le -2$, that $h(X) = -1$, but if this is not the case, then choice (B) could also be incorrect. But from what information we can see in the figure, the other choices are necessarily incorrect.