Sure. $a\equiv b\pmod{\mathfrak{P}}$ just means $a-b\in\mathfrak{P}$. Taking norms to any subfield $K$ of $\mathbb{Q}(\zeta_n)$ (e.g., $\mathbb{Q}$ or $\mathbb{Q}(\zeta_m)$) gives you $N_{\mathbb{Q}(\zeta_n)/K}(a-b)\in N_{\mathbb{Q(\zeta_n)}/K}\mathfrak{P}.$

For $K=\mathbb{Q}$, the latter norm is just $p^f$ where $f$ is the order of $p\pmod{n}$.

For $K=\mathbb{Q}(\zeta_m)$, the former norm is $(a-b)^{\phi(n)/\phi(m)}$ and the latter is $\mathfrak{p}^{f'}$, where $f'$ is the easily-calculated relative residue degree.

This doesn't give you an explicit congruence between $a$ and $b$, but given Gerry's answer, that might have been too much to ask for anyway. On the other hand, for small values of $\phi(n)/\phi(m)$, you get something at least slightly non-stupid out.

Sure. $a\equiv b\pmod{\mathfrak{P}}$ just means $a-b\in\mathfrak{P}$. Taking norms to any subfield $K$ of $\mathbb{Q}(\zeta_n)$ (e.g., $\mathbb{Q}$ or $\mathbb{Q}(\zeta_m)$) gives you $N_{\mathbb{Q}(\zeta_n)/K}(a-b)\in N_{\mathbb{Q(\zeta_n)}/K}\mathfrak{P}.$

For $K=\mathbb{Q}$, the latter norm is just $p^f$ where $f$ is the order of $p\pmod{n}$.

For $K=\mathbb{Q}(\zeta_m)$, the former norm is $(a-b)^{\phi(n)/\phi(m)}$ and the latter is $\mathfrak{p}^{f'}$, where $f'$ is the easily-calculated relative residue degree.

This doesn't give you an explicit congruence between $a$ and $b$, but given Gerry's answer, that might have been too much to ask for anyway. On the other hand, if $\phi(n)/\phi(m)$ is small or (as in Alex's answer) if $p$ has few factors in $\mathbb{Q}(\zeta_m)$, you get something at least slightly non-stupid out.

Sure. $a\equiv b\pmod{\mathfrak{P}}$ just means $a-b\in\mathfrak{P}$. Taking norms to any subfield $K$ of $\mathbb{Q}(\zeta_n)$ (e.g., $\mathbb{Q}$ or $\mathbb{Q}(\zeta_m)$) gives you $N_{\mathbb{Q}(\zeta_n)/K}(a-b)\in N_{\mathbb{Q(\zeta_n)}/K}\mathfrak{P}.$ For $K=\mathbb{Q}$, the latter norm is just $p^f$ where $f$ is the order of $p\pmod{n}$.

Sure. $a\equiv b\pmod{\mathfrak{P}}$ just means $a-b\in\mathfrak{P}$. Taking norms to any subfield $K$ of $\mathbb{Q}(\zeta_n)$ (e.g., $\mathbb{Q}$ or $\mathbb{Q}(\zeta_m)$) gives you $N_{\mathbb{Q}(\zeta_n)/K}(a-b)\in N_{\mathbb{Q(\zeta_n)}/K}\mathfrak{P}.$

For $K=\mathbb{Q}$, the latter norm is just $p^f$ where $f$ is the order of $p\pmod{n}$.

For $K=\mathbb{Q}(\zeta_m)$, the former norm is $(a-b)^{\phi(n)/\phi(m)}$ and the latter is $\mathfrak{p}^{f'}$, where $f'$ is the easily-calculated relative residue degree.

This doesn't give you an explicit congruence between $a$ and $b$, but given Gerry's answer, that might have been too much to ask for anyway. On the other hand, for small values of $\phi(n)/\phi(m)$, you get something at least slightly non-stupid out.