= Solution
Use the standard invariant <tabloid bilinear form>, with the <tabloid> <basis> orthonormal. Each permutation acts by an orthogonal <matrix>, so $\langle gx,y\rangle=\langle x,g^{-1}y\rangle$. Inversion permutes $C_t$ and preserves its signs. Therefore
$$
\boxed{\langle\kappa_tx,y\rangle=\sum_{g\in C_t}\operatorname{sgn}(g)\langle x,g^{-1}y\rangle=\langle x,\kappa_ty\rangle.}
$$
The <James submodule theorem> states that for any submodule $U\leq M^\lambda$, either $S^\lambda\subseteq U$ or $U\subseteq(S^\lambda)^\perp$. Use the allowed fact $\kappa_tM^\lambda=Fe_t$. The coefficient of $\{t\}$ in $e_t$ is one, so the displayed adjoint identity gives $\kappa_tx=\langle x,e_t\rangle e_t$. If $U$ is not orthogonal to $S^\lambda$, some $x\in U$ and some tableau $t$ have $\langle x,e_t\rangle\ne0$. Hence $e_t\in U$, and its permutation translates span $S^\lambda$. This proves the theorem.
Let $R=S^\lambda\cap(S^\lambda)^\perp$. Every proper submodule of $S^\lambda$ lies in $R$ by the theorem. If the restricted form is nonzero, $R$ is proper, hence is the unique maximal submodule and
$$
\boxed{D^\lambda=S^\lambda/R\text{ is the unique simple quotient}.}
$$
In the stronger nondegenerate case specified in the question, $R=0$, so $S^\lambda$ itself is simple. The nonzero-form version also covers the possibly degenerate case needed in the final calculation.
For shape $(n-1,1)$, identify $M$ with $F^n$ with <basis> $v_1,\ldots,v_n$. Its <Specht module> is the augmentation subspace $S=\{\sum a_iv_i:\sum a_i=0\}$, spanned freely by $v_i-v_n$, $1\leq i<n$. Its <Gram matrix> is $G=I_{n-1}+J_{n-1}$, with diagonal entries two and off-diagonal entries one. Equivalently, $S^\perp$ in $M$ is the span of $v_1+\cdots+v_n$. This vector belongs to $S$ exactly when $p\mid n$. Thus the restricted radical has <dimension> one if $p\mid n$ and zero otherwise, giving
$$
\boxed{\dim D^{(n-1,1)}=\begin{cases}n-2,&p\mid n,\\n-1,&p\nmid n.\end{cases}}
$$
For $n\geq3$ this quotient is nonzero and simple by the theorem. In the small exceptional case $n=2,p=2$, the canonical radical quotient is zero, as the same <dimension> formula states; it is then not a nonzero <simple module>. This qualification distinguishes the numerical formula from the nonzero-form hypothesis.
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