This function is defined as $k(f): B \times C \rightarrow A: (b,c) \mapsto k(f)(b,c) = f(c)(b).$
I already proved that this function is injective and surjective. Thus, it is bijective. But I don't know how to show that the function $k$ is also well-defined. Does anyone know how to do that? Thanks!
There really isn't much to check here. You have specified the value of $k(f)$ on each pair $(b,c)\in B\times C$, and thus defined a function $k(f)$ on $B\times C$. I guess maybe they want you to check that the value $f(c)(b)$ which is defined for $k(f)(b,c)$ actually exists and is an element of $A$. This is because $c\in C$ (so it is in the domain of $f$) and then $f(c)$ is a function $B\to A$ so $b$ is in its domain and the value $f(c)(b)$ is in $A$. So, $k(f)$ is indeed a function $B\times C\to A$, and since you have defined this for each $f\in (A^B)^C$, $k$ is a function $(A^B)^C\to A^{B\times C}$.
(Usually, when people talk about checking that a function is "well-defined", that means its definition involved using some non-unique representation of its inputs and you want to check that the output you get is independent of the representation chosen. There isn't anything like that going on here, though, so asking you to check the function is well-defined is a bit odd.)