$X$ is a separable, completely metrizable topological space equipped with its sigma algebra of Borel sets. $N$ is a countable space.
$X^N$ is the collection of all mappings from $N$ to $X$. It is equipped with product topology.
Is $N^X$ a Polish space?
Is $X^N$ a Polish space?
Is $X^X$ a Polish space?
It is obvious that $(2)$ is a Polish space, (3) is not in most cases. I guess $(1)$ is also Polish because $X^N$ and $N^X$ usually have the same structure?
If you really mean these spaces $A^B$ to consist of all functions $B\to A$, then (1) and (3) are too big to be Polish. They have cardinality $2^{\mathfrak c}$ where $\mathfrak c$ is the cardinal of the continuum. Separable metric spaces have cardinality at most $\mathfrak c$.
On the other hand, (2) is, like any product of countably many Polish spaces, well known to be Polish.
Let us work in a convenient category of topological spaces, ie assume that we have function spaces and sufficiently many nice spaces around. My favourite such category is the category of $\mathrm{QCB}_0$-spaces, which are $T_0$ quotients of countably-based spaces [1].
Inside the category of $\mathrm{QCB}_0$-spaces, Schröder has obtained a characterization of the Hausdorff spaces $\mathbf{X}$ such that $\mathbb{R}^\mathbf{X}$ is Polish [2]. These spaces are now called coPolish spaces. A Polish space is coPolish iff it is locally compact.
A space such as $2^{(\mathbb{N}^\mathbb{N})}$ even fails to be countably based.
[1] https://www.sciencedirect.com/science/article/pii/S0304397501001098
[2] https://onlinelibrary.wiley.com/doi/abs/10.1002/malq.200310111
