I am studying the first chapter of L. O. Arkeryd et. al: Nonstandard Analysis. Theory and Applications.
There it is shown that for the multiset $(\mathbb{X}, \mathcal{P}(\mathbb{X}))$ it is possible to find a nonstandard extension $(^*\mathbb{X}, {^*\mathcal{P}(\mathbb{X})})$ satisfying $\mathbb{X}\subset{^*\mathbb{X}}$ and ${^*\mathcal{P}(\mathbb{X})} \subset \mathcal{P}({^*\mathbb{X}})$. From thereon, it is always assumed that the considered nonstandard extension is of this form.
Recall that a subset of $^*\mathbb{X}$ is called internal if it is also an element of $^*\mathcal{P}(\mathbb{X})$.
In Exercise 5.5 one should show that every standard element of $^*\mathcal{P}(\mathbb{X})$ is an internal subset. This confuses me, since under the above assumption it looks like every element of $^*\mathcal{P}(\mathbb{X})$ is an internal subset. I also see how to show that every standard element of $^*\mathcal{P}(\mathbb{X})$ is an internal subset without the above assumption.
Basically my question is the following: Am I missing something here, i.e. under the assumption $^*\mathcal{P}(\mathbb{X})\subset \mathcal{P}(^*\mathbb{X})$ can there be an element of $^*\mathcal{P}(\mathbb{X})$ which is not an internal subset?