No, you cannot show this. For instance, it is consistent to have infinite Dedekind-finite sets whose power set is still Dedekind-finite. Now, if there is a surjection from $A$ to $\omega$, then there is an injection from $\omega$ (indeed, from $\mathcal P(\omega)$) to $\mathcal P(A)$, so $\mathcal P(A)$ is Dedekind-infinite.
Thus, if $\mathcal P(X)$ is infinite Dedekind-finite, then $X$ and $\omega$ are two non-empty sets with no bijections in either direction.
More generally, suppose that $X$ is not well-orderable, and let $\Theta$ be the first non-zero ordinal that $X$ does not surject onto. $\Theta$ clearly exists, since if $f:X\to\alpha$ is onto, then there is a partition of $X$ in type $\alpha$. The collection of partitions of $X$ is a set, and for any partition, there are only set-many orders we can put on its classes, so there are only set-many ordinals that work as $\alpha$.
Ok. Clearly, $\Theta$ does not surject into $X$, or $X$ would be well-orderable, since a subset of $\Theta$ would be in bijection with $X$ (by picking least representatives from each preimage). So $X$ and $\Theta$ are an example.