Let $p$ be an odd prime. Now $\mathbb{Q}_p (\sqrt{p})$ is a quadratic extension of $\mathbb{Q}_p$.
Is it correct to think that elements of $\mathbb{Q}_p (\sqrt{p})$ are of the form $a+\sqrt{p}b$ where $a,b \in \mathbb{Q}_p$?
If that is the case, then the norm of $a+\sqrt{p}b$ is $a^2-pb^2$?
This is true for any quadratic extension of any field $K$ of characteristic not $2$: if $L/K$ is a quadratic extension, then $L=K(\sqrt{d})$, its elements are of the form $a+b\sqrt{d}$, and their norm over $K$ is $a^2-db^2$.
This is because the Galois conjugate of $x=a+\sqrt{d}b$ is $\bar{x}=a-\sqrt{d}b$, so the norm is $$N_{L/K}(x)=x\bar{x}=(a+\sqrt{d}b)(a-\sqrt{d}b) = a^2-db^2.$$