A finite group is solvable if an only if it is $p$-solvable for every prime $p$. An example of a $p$-solvable group which is not solvable is the semidirect product $G = VSL(2,5)$, where $V$ is an elementary Abelian group of order $121$, and $SL(2,5)$ acts faithfully and irreducibly as a group of linear transformations on $V$. The group $G$ is $11$-solvable, but is not (Later remark: In fact, I think the above example is probably the example of minimal order of a finite group which is $p$-solvable, but not solvable, and does have order divisible by the prime $p$).
Later edit: Easier, but perhaps less interesting, is that a non-Abelian finite simple group $G$ is $p$-solvable if and only if the prime $p$ does not divide $|G|$. More generally, any finite group $G$ is $p$-solvable for all but finitely many primes $p$.
(Recall that a finite group $G$ is $p$-solvable for a prime $p$ if and only if every composition factor of $G$ is either of order $p$ or of order prime to $p$. It is in fact true that a finite group $G$ is solvable if and only if $G$ is $2$-solvable, but this requires the very deep Feit-Thompson Theorem that groups of odd order are solvable).