Let K be a finite field and let R,P be groups (with R a subgroup of P). I know that the irreducible KPmodules have dimensions 1,4 and 16 over K. I have a KPmodule M, and I know that M has dimension at most 5 over K. I also know that M does not have a quotient of dimension 1 over K. Moreover, if I consider M as a module over KR, it has a 2dimensional irreducible module. Is it necessarily the case that M is a 4dimensional irreducible module over KP? (my initial reasoning can be found in a comment below)

I see no reason in general if $\operatorname{Ext}_P^1(4,1) \neq 0$ that, on restriction to R the four dimensional simple can't drop down and have a 2dimensional submodule. I.e. On restriction to R it looks like:
If you knew on restriction to KR the only irreducible submodule was two dimensional then this would be ruled out. 

