= Solution
The implication is \b[false]. Choose
$$
\boxed{p=3,\quad k=2,\quad a=1,\quad m=4,\quad n=13,\quad\lambda=(12,1).}
$$
The inequalities $0\leq4<9$ and the relation $13=1\cdot9+4$ hold. At cell $(1,4)$, the <hook length> is $12-4+1=9=ap^k$.
For the <partition of an integer> $(12,1)$, the <Hook-length formula> gives
$$
\chi^{(12,1)}(1)=\frac{13!}{13\cdot11!}=12,
$$
which is divisible by $3$. This also agrees with the degree of the <standard representation of the symmetric group> $S_{13}$.
The obstruction is visible in the <core of a partition>: removing the horizontal <rim hook> of length nine leaves $\gamma=(3,1)$, which has no hook length nine and hence is the $9$-core. Its degree is $\chi^{(3,1)}(1)=3$. The large hook ensures the required <weight of a partition> here, but the core degree is not coprime to $p$.
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