 Solution

Take any $v\in V$. A smooth representation has an open stabilizer, so some sufficiently small congruence subgroup of $B$ fixes $v$. It can be written $K=T_m{U}_m$, where the diagonal entries in $T_m$ belong to $1+{\mathfrak{p}}^m$ and all entries above the diagonal in $U_m$ belong to ${\mathfrak{p}}^m$, with $m\ge 1$. These are compact-open in their respective groups, $T_m$ normalizes $U_m$, and their product is compact-open in $B$.

By part (a), this vector lies in $V^{K_i}$ for every sufficiently large $i$. Given any $u\in U$, the coordinate contraction $d^{-i}u{d}^i\to 1$ eventually puts it in $U_m$. Choose $i$ large enough for both properties. Then $u\in d^i{U}_m{d}^{-i}\subseteq K_i$, so $\rho (u)v=v$. Since $u$ and $v$ were arbitrary,Thus every admissible representation of this upper-triangular group factors through its diagonal quotient. The argument uses finite-dimensional fixed spaces essentially; smoothness alone does not force the unipotent action to be trivial.