To Add to Gerry's answer: to say that $a\equiv b \; ({\rm mod}\;{\mathfrak P})$ is equivalent to saying $a-b \in {\mathfrak P}$, so that ${\rm Tr}(a) - {\rm Tr}(b) \in \sum_{\sigma\in G}{\mathfrak P}^\sigma$, where $G$ is the Galois group of $\mathbb{Q}(\zeta_m)/\mathbb{Q}$. Since ${\mathfrak P}$ is prime, the sum is just the whole ring of integers, so this doesn't give you any information. On the other hand, it is true that ${\rm Norm}(a-b)$ is in $(p)$, but since the norm is not linear, I wouldn't expect this to tell you anything about ${\rm Norm}(a) - {\rm Norm}(b)$.

Edit: sometimes you do get the congruence you want for traces. See corrected answer: 

to say that $a\equiv b \; ({\rm mod}\;{\mathfrak P})$ is equivalent to saying $a-b \in {\mathfrak P}$, so that ${\rm Tr}(a) - {\rm Tr}(b) \in \sum_{\sigma\in G}{\mathfrak P}^\sigma$, where $G$ is the Galois group of $\mathbb{Q}(\zeta_m)/\mathbb{Q}$. If $p$ splits in $\mathbb{Q}(\zeta_m)$, then the sum is just the whole ring of integers, since ${\mathfrak P}$ is prime, hence maximal. So in this case, this doesn't give you any information. If on the other hand ${\mathfrak P}$ is the unique prime above $p$, then you get the desired statement that ${\rm Tr}(a-b)\in {\mathfrak P} \cap \mathbb{Q}$, so ${\rm Tr}(a)\equiv {\rm Tr}(b)\; {\rm mod}\; p$, as required.

As for the norm, it is true that ${\rm Norm}(a-b)$ is in $(p)$, but since the norm is not linear, I wouldn't expect this to tell you anything about ${\rm Norm}(a) - {\rm Norm}(b)$.