= Solution
Set $L=K_{\mathfrak P}$ and $F=k_{\mathfrak p}$, with valuation rings $S,R$. Let $e$ be the <ramification index>, $f$ the <residue-field degree>, and $p$ the <residue characteristic>. A base <uniformizer> $\pi$ generates $\mathfrak P^e$ in $S$, up to a unit. The free $R$-module $S$ has rank $ef$. For an integral $a\in S$, its <field trace> is the <matrix trace> of multiplication by $a$ on this module. Reduction modulo $\pi$ preserves the matrix trace.
The quotient $S/\pi S$ has a filtration
$$
S/\mathfrak P^e\supset\mathfrak P/\mathfrak P^e\supset\cdots\supset\mathfrak P^{e-1}/\mathfrak P^e\supset0.
$$
Each of the $e$ successive quotients is a one-dimensional vector space over $\kappa_L=S/\mathfrak P$. On each, multiplication by $a$ acts as multiplication by its residue $\bar a$, whose trace over $\kappa_F=R/\pi R$ is $\operatorname{Tr}_{\kappa_L/\kappa_F}(\bar a)$. A basis adapted to the filtration makes the multiplication matrix block triangular, so its trace is the sum of these quotient traces. This proves <trace reduction in a ramified local extension>:
$$
\overline{\operatorname{Tr}_{L/F}(a)}=e\operatorname{Tr}_{\kappa_L/\kappa_F}(\bar a).
$$
Since the hypothesis says $p\mid e$, the right side is zero in $\kappa_F$. Thus
$$
\boxed{\operatorname{Tr}_{K_{\mathfrak P}/k_{\mathfrak p}}(a)\in\mathfrak p\mathcal O_{k_{\mathfrak p}}.}
$$
No normality hypothesis is needed. The original PDF only requires divisibility of $e$, not divisibility exactly once; the converted TeX has corrupted that condition and several local subscripts.
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