Let $\Omega$ be an open set in $\mathbb{R}^n$ and $E$ a Borel set. The relative perimeter of $E$ w.r.t. $\Omega$ is defined as $$P(E,\Omega)=\sup\left\{\int_{\Omega}\chi_E(x) \mathrm{div}\boldsymbol{\phi}(x) \, \mathrm{d}x : \boldsymbol{\phi}\in C_c^1(\Omega,\mathbb{R}^n),\ \|\boldsymbol{\phi}\|_{L^\infty(\Omega)}\le 1\right\},$$ see here https://en.wikipedia.org/wiki/Caccioppoli_set.
Suppose that $E\cap \Omega$ is non-empty. Is $P(E,\Omega)=P(E\cap \Omega,\mathbb{R}^n)$? Or is $P(E,\Omega)=P(E\cap \Omega,\mathbb{R}^n)$+some (n-1)-dimensional Hausdorff measure?
It is $$P(E\cap \Omega,\mathbb{R}^n)=\sup\left\{\int_{\mathbb{R}^n}\chi_{E\cap\Omega}(x) \mathrm{div}\boldsymbol{\psi}(x) \, \mathrm{d}x : \boldsymbol{\psi}\in C_c^1(\mathbb{R}^n,\mathbb{R}^n),\ \|\boldsymbol{\phi}\|_{L^\infty(\mathbb{R}^n)}\le 1\right\}$$ $$=\sup\left\{\int_{\mathbb{R}^n}\chi_{E}\chi_{\Omega}(x) \mathrm{div}\boldsymbol{\psi}(x) \, \mathrm{d}x : \boldsymbol{\psi}\in C_c^1(\mathbb{R}^n,\mathbb{R}^n),\ \|\boldsymbol{\phi}\|_{L^\infty(\mathbb{R}^n)}\le 1\right\}$$ $$=\sup\left\{\int_{\Omega}\chi_{E}(x) \mathrm{div}\boldsymbol{\psi}(x) \, \mathrm{d}x : \boldsymbol{\psi}\in C_c^1(\mathbb{R}^n,\mathbb{R}^n),\ \|\boldsymbol{\phi}\|_{L^\infty(\mathbb{R}^n)}\le 1\right\}$$ From there I conclude $P(E,\Omega)\le P(E\cap \Omega,\mathbb{R}^n)$. However, does $P(E,\Omega)\ge P(E\cap \Omega,\mathbb{R}^n)$ hold? Or how do the perimeters relate?
They aren't the same. It helps to use sets with elementary geometry so that "perimeter" has its usual meaning.
As an example, let $\Omega$ be the open unit disk in $\mathbb R^2$ and let $E$ be the right-half of $\Omega$: $$E = \{(x,y) : x^2 + y^2 < 1,\ x \ge 0\}.$$
The portion of the boundary of $E$ inside $\Omega$ is a vertical segment with length $2$ so that $P(E,\Omega) = 2$.
On the other hand, the boundary of $E$ consists of that vertical line segment along with a semicircle. Since $E \cap \Omega = E$ you get $P(E \cap \Omega) = P(E) = 2 + \pi$.