Thank you for your solution.
Here is another solution that \(\displaystyle{f}\) is one to one according to the hypothesis
that \(\displaystyle{\left\{A,B\right\}}\) is a partition of \(\displaystyle{E}\), that is
\(\displaystyle{E=A\cup B\,\,,A\cap B=\varnothing}\) . Let \(\displaystyle{X\,,Y\in\mathbb{P}(E)}\) such
that \(\displaystyle{f(X)=f(Y)}\). Then, \(\displaystyle{X\cap A=Y\cap A\,\,\,,X\cap B=Y\cap B}\).
We'll prove that \(\displaystyle{X=Y}\) . For this purpose, let \(\displaystyle{x\in X}\). Then, \(\displaystyle{x\in E}\)
and \(\displaystyle{x\in A}\) or \(\displaystyle{x\in B}\) . If \(\displaystyle{x\in A}\), then :
\(\displaystyle{x\in X\cap A\implies x\in Y\cap A\implies x\in Y\,\land x\in A}\), so : \(\displaystyle{x\in Y}\) . If \(\displaystyle{x\in B}\)
then \(\displaystyle{x\in X\cap B\implies x\in Y\cap B\implies x\in Y}\). In any case, \(\displaystyle{X\subseteq Y}\).
Similarly, \(\displaystyle{Y\subseteq X}\) and finally, \(\displaystyle{X=Y}\) .
Here is another solution that \(\displaystyle{A\cup B=E}\)according to the hypothesis that \(\displaystyle{f}\) is
one to one an onto \(\displaystyle{\mathbb{P}(A)\times \mathbb{P}(B)}\) .
Suppose that \(\displaystyle{A\cup B\neq E}\). Then, there exists \(\displaystyle{x\in E}\) such that
\(\displaystyle{x\notin A\,\,,x\notin B}\). Therefore,
\(\displaystyle{f(\left\{x\right\})=\left(\left\{x\right\}\cap A,\left\{x\right\}\cap B\right)=\left(\varnothing,\varnothing\right)=f\,(\varnothing)}\)
and since the function \(\displaystyle{f}\) is one to one, we get \(\displaystyle{\left\{x\right\}=\varnothing}\), a contradiction,
so \(\displaystyle{A\cup B=E}\) .
