Plancharel and absolute values

Question

Let $h\in C_0^\infty(\mathbb{R})$, that is, $h$ is a smooth and compactly supported real-valued function. Assume further that $\Vert h\Vert_{L^\infty}+\Vert h'\Vert_{L^\infty}\leq 1$. Now, consider $\mathcal{P}_h: L^2\times L^2\to L^2$ to be the operator given in terms of its Fourier transform $$ \qquad \qquad \widehat{\mathcal{P}_h(f,g)}(\xi)= \int_\mathbb{R}\big(h(\xi)-h(y)\big)\hat{f}(y)\widehat{g}(\xi-y)dy. \qquad \qquad (*) $$ Here, notation $\hat{\cdot}$ stands for the Fourier transform. Now, by the Mean Value Theorem we know that there exists a $\xi^\star\in(\min\{\xi,y\},\max\{\xi,y\})$ such that $$ h(\xi)-h(y)=(\xi-y)h'(\xi^\star). $$ Now here is my problem: I would like to say that, for any $f,g\in L^2$, with $g'\in L^2$ and $p\in L^\infty$ we have $$ \qquad \qquad \left\vert\int_{\mathbb{R}}\mathcal{P}_h(f,g)(x)p(x)dx\right\vert\leq C\Vert f\Vert_{L^2}\Vert g'\Vert_{L^2}\Vert p\Vert_{L^\infty}, \qquad \qquad (**) $$ where, of course, the constant depends on $h$. The idea for me (if I'm not wrong) would be to use this MVT on $h$, to pass the factor $(\xi-y)$ as a derivative of $g$. However, I have the feeling that I cannot just use Plancharel and then do Cauchy-Schwarz/Holder, specifically because $\xi^\star$ depends on both $\xi$ and $y$, and hence I am not sure if I can just put it in only one of the $L^2$ norms. On the other hand, I thought just to use Plancharel, and then to bound $h'(\xi^\star)$ just by $1$, but this requires to put some absolute values inside the integral, with would make impossible to go back to $x$ by Plancharel again, right? So I am not sure how to prove something like $(**)$ (if it is true). I hope the question is clear enough. Any ideas?