Let $\Omega$ be a bounded Lipschitz domain in $\mathbb{R}^d$ and consider the following situation. $$\phi(x) = \int_{\partial \Omega} \frac{x-y}{|x-y|^d} f(y) d\sigma(y)$$ Where $\partial \Omega$ denotes the boundary of $\Omega$ and $\sigma$ denotes the surface measure. I want to show that the nontangential maximal function of $\phi$ is bounded on $L^p(\partial \Omega)$ when $f \in L^p(\partial \Omega)$ for $1 < p < \infty$. To do so we look at the definition of the nontangential maximal function (denoted by $(\phi)^{*})$. Let $P \in \partial \Omega$ denote a point on the boundary. Then $$ (\phi)^*(P) = \sup_{\substack{x \in \Omega \\ |x-P| < C_1 \operatorname{dist}(x,\partial \Omega)}} \left| \int_{\partial \Omega} \frac{x-y}{|x-y|^d} f(y) d\sigma(y) \right|$$ I have split the integral in multiple parts and managed to bound some of them by the maximal function of $f$ in $P$. However I'm missing one part.
I want to show that $$ \sup_{\substack{x \in \Omega \\ |x-P| < C_1\operatorname{dist}(x,\partial \Omega)}} \left| \int_{|P-y| > 2|P-x|} \frac{|P-y|}{|P-y|^d} f(y) d\sigma(y) \right| \leq C_2 \mathcal{M}_{\partial \Omega}(f)(P)$$ With the maximal function defined as $$\mathcal{M}_{\partial \Omega}(f)(P) = \sup_{(B\cap \partial \Omega) \ni P}\frac{1}{\sigma\{B\cap \partial \Omega\}}\int_{B \cap \partial \Omega} |f(y)| d\sigma(y)$$