Maximal subsets of a point set which fit in a unit disk
Let's define a function $f: \Bbb R \times \Bbb R \rightarrow \left [0, n \right ] $, returning a number of points from the set $P$, covered by a unit disk with the center at the point $(x, y)$. This is a piecewise constant function, and it's easy to see that its domain can be thought as a planar subdivision, defined by all intersections of unit disks, centered at points from the set $P$. This subdivision contains vertices (= intersection points), edges (= circular arcs) and faces (= pieces of the plane, where the function $f$ returns the same value). We'll say that these faces are labeled by this value.
We'll assume that the planar graph, defined by this subdivision, is a 4-regular one - it's a common assumption, meaning that all the points in $P$ are in general position (each intersection point belongs to exactly two circles). An example of such subdivision for $n=3$ is below, where the label for each face (including the external one) is shown in red color.

As far as I understand your maximal subsets $P_i$ can be associated with such faces of this subdivision, which have labels larger than all labels of their neighboring faces. It's kind of static interpretation of your &quot;moving disk&quot; semantics, which you used in your definition of the maximal subset.
The subdivision will contain as many faces as possible in the case when all the unit circles pairwise intersect each other. It can be shown, that in this case:

number of vertices is $n(n-1)$
number of edges is $2n(n-1)$
number of faces is $n(n-1)+2$

So, answers to your questions will be:

No, the number $k$ of maximal subsets is $O(n^2)$, because it can't be more than the number of all faces.
Yes, all the maximal subsets can be found in $O(n^3)$ time by naive scanning algorithm.

As for name of this problem - it looks like it may be related to some variations of the Unit Disk Cover problem.