Solution:

The answer is $(\mathbf{C})$. The only one who is right is Carlo.

Alberto is wrong: indeed, if there existed $X, Y, Z, P$ belonging to the same plane such that $P X, P Y$ and $P Z$ are pairwise perpendicular, then $P Z \perp P X, P Z \perp P Y$ and hence $P Y \parallel P X$, which is absurd.

Barbara is wrong: 4 such points are given, for example, by the endpoints of 3 edges of a cube sharing a common vertex (which will be $P$), or more generally by the vertices of any right-angled tetrahedron.

Carlo is right: indeed, since $X Y^{2} = P X^{2} + P Y^{2} < P X^{2} + P Y^{2} + 2 \cdot P Z^{2} = X Z^{2} + Y Z^{2}$, then $\widehat{X Z Y}$ is acute, and similarly $\widehat{Z X Y}$ and $\widehat{X Y Z}$ are also acute. An alternative argument to show that Carlo is right is the following: consider the sphere with diameter $X Y$, then $P$ belongs to this sphere and the line $P Z$ is perpendicular to the plane on which $P, X, Y$ lie, hence every point of this line (except $P$) is outside the sphere. But then $\widehat{X Z Y}$ is acute (and similarly $\widehat{Z X Y}$ and $\widehat{X Y Z}$ are also acute).

Daria is wrong: by what was said above, it is necessary that $X Y Z$ be acute-angled, so $X Y Z$ cannot be right-angled; in particular, it is not sufficient that $X Y Z$ not be obtuse-angled.