= Solution
Put
$$
W=\operatorname{Ind}_T^B\theta.
$$
Since $B=U\rtimes T$ and $U\cap T=1$, restriction to $U$ gives its regular representation:
$$
\operatorname{Res}_U^BW\cong\mathbb C[U].
$$
As $U\cong(\mathbb F_p,+)$ is abelian, this is the direct sum of all $p$ one-dimensional characters of $U$, each with multiplicity one.
The diagonal subgroup $T$ permutes these one-dimensional weight spaces. It fixes the trivial character, while part (ii) shows that the nontrivial characters split into two orbits, indexed by the nonzero squares and nonsquares. Let $W_0,W_+,W_-$ be the sums of the weight spaces in these three orbits. Then
$$
W=W_0\oplus W_+\oplus W_-
$$
as $B$-representations, and
$$
\dim W_0=1,
\qquad
\dim W_+=\dim W_-=\frac{p-1}{2}.
$$
Each summand is irreducible. Any nonzero $B$-subrepresentation decomposes into $U$-weight spaces. If it contains one of the one-dimensional weight spaces in an orbit, transitivity of the $T$-action forces it to contain every weight space in that orbit, hence the whole corresponding $W_0$, $W_+$, or $W_-$. The three summands are pairwise nonisomorphic because their restrictions to $U$ have disjoint character supports. Therefore the <induction from the diagonal subgroup of upper-triangular SL2> decomposition has the required three irreducible constituents of dimensions
$$
\boxed{1,\quad\frac{p-1}{2},\quad\frac{p-1}{2}}.
$$
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