Bijection between $2^{\mathbb R}$ and $\mathbb{ R ^ R}$

Once you have the bijection $P:2^\mathbb{N}\cong\mathbb{R}$, you can build your desired bijection as follows.

First, note that $\mathbb{N}\times 2^\mathbb{N}\cong 2^\mathbb{N}$, essentially by the bijection that associates $(n,A)$ with $\{n\}\cup (n+1+A)$, except that this will miss the empty set on the right, but we can fix this by composing with a map witnessing $2^\mathbb{N}-\{\varnothing\}\cong 2^\mathbb{N}$, such as shifting on a fixed countable subset.

Now simply observe that $$\mathbb{R}^\mathbb{R}\cong (2^\mathbb{N})^{(2^\mathbb{N})}\cong 2^{(\mathbb{N}\times 2^\mathbb{N})} \cong 2^{(2^\mathbb{N})} \cong 2^\mathbb{R}, $$

which provides the desired bijection. The first map is conjugation by $P$, simply composing with $P^{-1}$ and $P$ before and after; the second map is an easy exercise in parenthesis rearranging; the third map applies the observation above; and the final map applies $P$.