I don't think you mean $d = \frac{q-1}{p}$ in your presentation. I think (as Ben Webster has already implicitly noted) you want $yxy^{-1} = x^{b}$ where $b^{p} \equiv 1 \not \equiv b$ (mod $q$). Ben's answer is correct, of course, but I'll try to outline a solution at the level that students who've just seen Maschke's theorem (and maybe Schur's Lemma) could see without much further theory. So, irreducible complex characters of Abelian groups have degree $1$ by Schur's Lemma. There are $p$ irreducible characters of $G$ of degree $1$ with $\langle x \rangle$ in their kernel. Let's show directly that an irreducible representation of $G$ which does not contain $x$ in its kernel has degree at least $p.$ (Then a dimension count forces all non-linear irreducible characters to have degree exactly $p$). This amounts to explicitly constructing directly the induced modules Ben discusses, but assuming no extra theory a priori. Let $V$ be the associated module. There must some non-zero vector $v \in V$ and some $\omega \neq 1$ such that $vx = \omega v.$ Then $v$ must be a primitive $q$-th root of unity as $x^{q} = 1.$ Now $vyxy^{-1} = vx^{b} = \omega^{b}v.$ Hence $vy$ is an eigenvector of $x$ with eigenvalue $\omega^{b}.$ Repeating the argument several times, we see that $v,vy,vy^{2},\ldots vy^{p-1}$ are eigenvectors of $x$ such that $vy^{j}$ is associated to eigenvalue $\omega^{b^{j}}.$ Hence $V$ has dimension at least $p,$ as claimed.

I don't think you mean $d = \frac{q-1}{p}$ in your presentation. I think (as Ben Webster has already implicitly noted) you want $yxy^{-1} = x^{b}$ where $b^{p} \equiv 1 \not \equiv b$ (mod $q$). Ben's answer is correct, of course, but I'll try to outline a solution at the level that students who've just seen Maschke's theorem (and maybe Schur's Lemma) could see without much further theory. So, irreducible complex characters of Abelian groups have degree $1$ by Schur's Lemma. There are $p$ irreducible characters of $G$ of degree $1$ with $\langle x \rangle$ in their kernel. Let's show directly that an irreducible representation of $G$ which does not contain $x$ in its kernel has degree at least $p.$ (Then a dimension count forces all non-linear irreducible characters to have degree exactly $p$). This amounts to explicitly constructing directly the induced modules Ben discusses, but assuming no extra theory a priori. Let $V$ be the associated module. There must some non-zero vector $v \in V$ and some $\omega \neq 1$ such that $vx = \omega v.$ Then $\omega$ must be a primitive $q$-th root of unity as $x^{q} = 1.$ Now $vyxy^{-1} = vx^{b} = \omega^{b}v.$ Hence $vy$ is an eigenvector of $x$ with eigenvalue $\omega^{b}.$ Repeating the argument several times, we see that $v,vy,vy^{2},\ldots vy^{p-1}$ are eigenvectors of $x$ such that $vy^{j}$ is associated to eigenvalue $\omega^{b^{j}}.$ Hence $V$ has dimension at least $p,$ as claimed.