= Solution
If some $a_j\geq p$, then $a_j!=0$ in $\mathbb F$. Part b(ii) makes every pairing $\langle e(t),e(u)\rangle$ zero, so the <Tabloid bilinear form> vanishes identically on $S^\lambda$ and $S^\lambda\subseteq(S^\lambda)^\perp$.
Conversely, if $\lambda$ is $p$-regular, every $a_j!$ is nonzero. Fact 1 supplies tableaux $t,u$ with
$$
\langle e(t),e(u)\rangle=\prod_{j=1}^n(a_j!)^j\ne0.
$$
Thus the restriction of the form is not identically zero and $S^\lambda\not\subseteq(S^\lambda)^\perp$.
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