Notation:
$\Omega$ denotes a $\sigma-$Field
$X$ is a function onto $\Omega$
$X^{-1}$ is pre-image
$P(\cdot)$ is a probability measure
$X^{-1}$ is also a $\sigma-$Field, as a consequence of properties of preimage
Claim:
$P(\{ X^{-1}(x)\})=p$ $\;\forall x \in \Omega$
Then $\Omega$ has finite number of points
Logic:
By Definition of PDF: $k \cdot \sum P(\{ X^{-1}(x)\})=1$
Thus, this sum should be definite, as in $k$ should be finite, hence the preimage should be finite ,ie, $\Omega$ should have finite number of points.
Is This reasoning correct?
Edit:
I don't think the original question says anything about probability distributions.
So considering $P$ as only a probability measure:
$\cup A_i = \Omega$ for some $A_i$
By Kolmogorov additivity axiom: $P(X^{-1}( \cup A_i)) = \sum P(X^{-1}(A_i))$
$= P(X^{-1}(\Omega)) = 1$
Hence number of elements of $\Omega$ is $n \in \mathbb{N}$ such that $n\cdot p=1$