Conceptually P=3 would belong here as well: its cells then would be tets only and its vertex figure would be an ike. Because this series generally describes convex polytopes, that one clearly becomes nothing but ex. However, because the circumradius of ike is smaller than 1, this special case happens to belong to spherical geometry rather than to the hyperbolic one, i.e. asks fon non-infinite matrix values.

Even more degenerate would become the case P=2. There the vertex figure would become an oct. However the second cell type then would become degenerate, in fact just subdimensional triangles, i.e. alike to the other already existent faces. Thence those "cells" no longer would count as cells, as the below incidence matrix indicates, they then would belong to the other block of faces instead. That special case then would describe nothing but hex.

Incidence matrix according to symbol extension

x3o:s3sPs   (N,M → ∞)
            q = LCM(2,3,P)/6P

  N |  6PqM |  9PqM  6PqM |  8PqM 6Pq  verf: s3sPs
----+-------+-------------+----------
  2 | 3PqNM |     3     2 |     4   1
----+-------+-------------+----------
  3 |     3 | 3PqNM     * |     2   0
  3 |     3 |     * 2PqNM |     1   1
----+-------+-------------+----------
  4 |     6 |     3     1 | 2PqNM   *  tet
6qM |  3PqM |     0  2PqM |     *   N  x3oPo