For every non-trivial irreducible character $\chi$ of a finite $p$-group $P,$ we may choose an element $z in P$ such that $\chi(z) = \chi(1) \omega$ for some primitive $p$-th root of unity $\omega$ (this is an easy consequence of Schur's Lemma and the fact that the image of $P$ in the associated representation has non-trivial center). Now let $\zeta$ be a primitive $p^{e}$-th root of unity, where $P$ has exponent $p^{e}.$ Then ${\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q}]$${\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ acts on the irreducible characters of $P,$ and it is clear that $\chi$ is in an orbit of length divisible by $p-1.$ Since $\chi$ was an arbitrary non-trivial irreducible character, and since all irreducible characters in the same orbit have the same degree it follows both that the total number of non-trivial irreducible characters, and the total number of irreducible non-linear characters of $P$ are multiples of $p-1$ (in fact, the number of irreducible characters of $p$ of any fixed degree $p^{d} >1$ is a multiple of $p-1.$ And yes, a similar statement holds for conjugacy classes. If $C$ is a non-trivial conjugacy class of $P$, say containing an element $x,$ then the length of the conjugacy class only depends on $\langle x \rangle$. If $x$ has order $p^{e},$ then $\langle x \rangle$ has $p^{e-1}(p-1)$ generators, but if $y$ is another generator, and $y^{p^{e-1}} \neq x^{p^{e-1}},$then $x$ and $y$ are not conjugate within $p.$ Since $\langle x \rangle$ contains $p-1$ elements of order $p,$ it follows that the number of conjugacy classses of non-trivial elements of $P$ which have length $[P:C_{P}(x)]$ is divisible by $p-1.$ In particular, the number of conjugacy classes of length greater than $1$ is divisible by $p-1.$
