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Let $Z$ be the center of $G={\rm GL}(2,{\mathbb F}_q )$ and $$ M=\left\{ \left( \begin{array}{cc} a & b \\ 0 & 1\end{array}\right)\in G\right\} $$ be the mirabolic subgroup of $G$, so that $B=ZM$. Fix a non trivial character $\psi$ of $U$. By the theory of Kirillov models, the representation $\mu ={\rm Ind}_U^M \, \psi$ is irreducible (of dimension $q-1$), and for any irreducible representation $\pi$ of $G$, of dimension $>1$, we have $$ {\rm dim}\, {\rm Hom}_M (\pi ,\mu ) =1 $$ If $\pi$ is cuspidal, then ${\rm dim}\, \pi =q-1$, so that, $\pi_{\mid M}\simeq \mu$. So the restriction of $\pi$ to $M$ does not depends on the irreducible cuspidal representation $\pi$. The restriction of those $\pi$ to $B$ only depends on the central character $\omega_\pi$ of $\pi$.