= 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 $K=T_mU_m$, 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 $m\ge1$. 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 $d^{-i}ud^i\to1$ eventually puts it in $U_m$. Choose $i$ large enough for both properties. Then $u\in d^iU_md^{-i}\subseteq K_i$, so $\rho(u)v=v$. Since $u$ and $v$ were arbitrary,
$$
\boxed{\rho(u)=\operatorname{id}_V\quad\text{for every }u\in U.}
$$
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.
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