It seems that the tool being used is the Minkowski integral inequality , with Proposition 3.6 of Duoandikoetxea.
Minkowski Integral inequality (Duoandikoetxea, Fourier Analysis,xviii (Preliminaries)) : Let $(X,\mu)$ and $(Y,\nu)$ be $\sigma$-finite measure spaces. For all appropriate $f(x,y) : X \times Y \to \mathbb R$ (that is ,for which the RHS of the inequality exists),$$
\left(\int_X \left|\int_{Y} f(x,y)d\nu(y)\right|^p d \mu(x)\right)^{\frac 1p} \leq \int_Y \left(\int_{X}|f(x,y)|^pd \mu(x)\right)^{\frac 1p} d \nu(y)
$$
we write this as
$$
\int_X \left|\int_{Y} f(x,y)d\nu(y)\right|^p d \mu(x) \leq \left(\int_Y \left(\int_{X}|f(x,y)|^pd \mu(x)\right)^{\frac 1p} d \nu(y)\right)^{p}
$$
This can be thought of as an integral analogue of the triangle inequality for $L^p$ functions.
At this point, we cannot quite apply this exactly to our situation, because the measure given by $m'(t)dt$ is a signed measure (where some sets may have a negative measure) rather than a conventional measure. However, note that one of the assumptions on $m$ usually imposed is that it is of bounded variation (it's a consequence of the fact that $\|m'\|_{L^1}$ exists).
In that case, we have the functions $h_1(t)= m'(t)1_{m'(t) \geq 0}$ and $h_2(t) = m'(t)1_{m'(t) \leq 0}$. Call the measures $dm_1(t) = h_1(t)dt$ and $dm_2(t)= -h_2(t)dt$. It is clear that $\int_A m'(t)dt = m_1(A)-m_2(A)$ and $\int_A |m'(t)|dt = m_1(A) + m_2(A)$ for any measurable set $A$. This is a Jordan decomposition of $m'(t)dt$ that we've performed : getting two measures on which we'll execute the Minkowski integral inequality.
The way to start is to somehow relate the LHS with $d \nu(y) = m'(t)dt$ and separately with $dm_1$ and $dm_2$. For this, note that $$
\int_{A} m'(t)dt = m_1(A)-m_2(A) \implies \left|\int_{A} m'(t)dt\right| \leq |m_1(A)| + |m_2(A)|
$$
and therefore $$
\left|\int_{A} m'(t)dt\right|^p \leq \left(|m_1(A)| + |m_2(A)|\right)^p \leq 2^p (|m_1(A)|^p + |m_2(A)|^p)
$$
for any measurable set $A$. The last inequality follows from two inequalities $$
x +y \leq 2\max(x,y) \quad ; \quad \max(x^p,y^p) \leq (x^p+y^p)
$$
for all $x,y$ positive. Once we see this, we can split any $L^p$-integrable function into a positive-negative part , approximate each part by indicators, and obtain for any $m_1(t)dt$ $L^p$-integrable $g$ that $$
\left|\int_{\mathbb R} g(t)m'(t)dt\right|^p \leq 2^p \left(\left|\int_{\Bbb R} g(t)dm_1(t)\right|^p + \left|\int_{\Bbb R} g(t)dm_2(t)\right|^p\right)
$$
We get by using $g(t) = \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)$ for any fixed $\xi$ that $$
\left|\int_{\mathbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^p \leq 2^p \left(\left|\int_{\Bbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)dm_1(t)\right|^p + \left|\int_{\Bbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)dm_2(t)\right|^p\right)
$$
Integrating w.r.t $d\xi$, we have
$$
\int_{\mathbb R} \left|\int_{\mathbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^pd \xi\\ \leq 2^p\left(\int_{\mathbb R}\left|\int_{\Bbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)dm_1(t)\right|^p d \xi + \int_{\mathbb R}\left|\int_{\Bbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)dm_2(t)\right|^p d \xi\right) \tag{1}
$$
To apply the Minkowski integral inequality, we have for $i=1,2$, $$
\int_{\mathbb{R}}\left|\int_\mathbb{R}\underbrace{{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)}_{f(\xi,t)}\underbrace{dm_i(t)}_{d \nu(t)}\right|^p\underbrace{d\xi}_{d \mu(\xi)}
$$
It's obvious that $d \xi$ is a sigma-finite measure on $\mathbb R$. We also have $\|m_i\|_{L^1}\leq \|m'\|_{L^1}$ so the $m_i$ are finite measures. Finally, assuming that the RHS of the integral inequality will be finite, we'll write the expression for it :
$$
\int_{\mathbb{R}}\left|\int_\mathbb{R}\underbrace{{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)}_{f(\xi,t)}\underbrace{dm_i(t)}_{d \nu(t)}\right|^p\underbrace{d\xi}_{d \mu(\xi)} \leq \left(\int_{\Bbb R}\left(\int_{\Bbb R}|{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)|^p d\xi\right)^{\frac 1p} dm_i(t)\right)^{p} \tag{2}
$$
We now use proposition $3.6$ of the book. (Page 59). Recall the operator $S_{a,b}$ defined in the book. It is fairly clear to see that $$
\mathcal F(S_{(t,\infty)}f) (\xi) = (\chi_{(t,\infty)}\hat{f})(\xi)
$$
since $S_{a,b}$ is the operator associated with the multiplier $\chi_{a,b}$. Inverting the Fourier transform, $$
S_{(t,\infty)}f = \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})
$$
However, this means that we can write the RHS of $(2)$ as
$$
\left(\int_{\Bbb R}\left(\int_{\Bbb R}|{\mathcal{F}^{-1}}(\chi_{(t,\infty)}\hat{f})(\xi)|^p d\xi\right)^{\frac 1p} dm_i(t) \right)^p = \left(\int_{\mathbb R} \|S_{(t,\infty)}\|_p dm_i(t)\right)^p
$$
Using proposition $3.6$, there exists a constant $C_p$ independent of $t$ such that $$
\|S_{(t,\infty)}\|_p \leq C_p \|f\|_p\quad \forall -\infty \leq t \leq \infty
$$
Substituting this above and taking the constant $C_p\|f\|_p$ out of the integral gives $$
\left(\int_{\mathbb R} \|S_{(t,\infty)}\|_p dm_i(t)\right)^p \leq \left(C_p \|f\|_p \int_{\mathbb R} dm_i(t)\right)^p \leq C_p^p \|f\|^p_p\|m'\|^p_1
$$
For $i=1,2$.Combining $2$ and $1$ now gives us $$
\int_{\mathbb R} \left|\int_{\mathbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^pd \xi \leq 2^p\left(2C_p^p \|f\|^p_p\|m'\|^p_1\right) \leq 2^{p+1} C_p^p \|f\|_p^p \|m'\|_1^p
$$
Taking the $p$th root gives $$
\left(\int_{\mathbb R} \left|\int_{\mathbb R} \mathcal F^{-1}(\chi_{(t,\infty)}\hat{f})(\xi)m'(t)dt\right|^pd \xi \right)^{\frac 1p} \leq 2^{\frac{p+1}p}C_p \|f\|_p\|m'\|_1
$$
where the constant term depends only on $p$, as desired.
For each $f\in C_c^\infty(\mathbb R^n)$, we have
$$\left(T_{m*\psi}f\right)\hat{}(\xi)=(m*\psi)(\xi)\hat f(\xi)=\int_{\mathbb R^n}m(\xi-\eta)\psi(\eta)\hat f(\xi)\,d\eta.$$
Hence
\begin{align*}
\left(T_{m*\psi}f\right)(x)&=\int_{\mathbb R^n}\int_{\mathbb R^n}m(\xi-\eta)\psi(\eta)\hat f(\xi)\,d\eta e^{2\pi ix\cdot \xi}\,d\xi\\
&=\int_{\mathbb R^n}\left(\int_{\mathbb R^n}m(\xi-\eta)\hat f(\xi)e^{2\pi ix\cdot \xi}\,d\xi\right)\psi(\eta)\,d\eta.
\end{align*}
Now we calculate the integral in the bracket. We define two operators to simplify our notations:
$$M_af(x)=e^{2\pi ia\cdot x}f(x),\qquad \tau_af(\xi)=f(\xi-a),\qquad a\in\mathbb R^n.$$
Then clearly we have $(M_af)\hat{ }=\tau_a(\hat f)$. Now,
\begin{align*}
\int_{\mathbb R^n}m(\xi-\eta)\hat f(\xi)e^{2\pi ix\cdot \xi}\,d\xi&=\int_{\mathbb R^n}m(\xi-\eta)\int_{\mathbb{R}^n} f(y)e^{-2\pi iy\cdot\xi}\,dy e^{2\pi ix\cdot \xi}\,d\xi\\
&=\int_{\mathbb R^n}\left(\int_{\mathbb R^n}m(\xi-\eta)e^{2\pi i(x-y)\cdot(\xi-\eta)}\,d\xi\right)e^{2\pi i(x-y)\cdot\eta}f(y)\,dy\\
&=\int_{\mathbb R^n}\overset{\vee}{m}(x-y)e^{2\pi i(x-y)\cdot\eta}f(y)\,dy\\
&=\left(\left(M_\eta \overset{\vee}{m}\right)*f\right)(x).
\end{align*}
We claim that
$$\left\|\left(M_a \overset{\vee}{m}\right)*f\right\|_{L^p}\leq \left\|T_{m}\right\|_{L^p\to L^p}\|f\|_{L^p}\qquad\forall a\in\mathbb R^n.\tag{$*$}$$
Assuming the claim, by Minkowski's inequality we get
$$\left\|T_{m*\psi}f\right\|_{L^p}\leq\int_{\mathbb R^n} |\psi(\eta)|\left\|\left(M_\eta \overset{\vee}{m}\right)*f\right\|_{L^p}\,d\eta\leq \left\|T_{m}\right\|_{L^p\to L^p}\|\psi\|_{L^1}\|f\|_{L^p}.$$
This gives the desired result.
Finally, we prove the claimed identity $(*)$. Since $\left(M_a \overset{\vee}{m}\right)\hat{}=\tau_a m$, we have $\left(M_a \overset{\vee}{m}\right)*f=T_{\tau_am}f$. It is easy to check that
$$T_{\tau_am}f= M_a\left(T_m\left(M_{-a}f\right)\right).$$
By definition of $M_a$, we have $|M_af|=|f|$, so $\|M_af\|_{L^p}=\|f\|_{L^p}$. Therefore,
$$\left\|T_{\tau_am}f\right\|_{L^p}=\left\|T_m\left(M_{-a}f\right)\right\|_{L^p}\leq\left\|T_{m}\right\|_{L^p\to L^p}\|M_{-a}f\|_{L^p}=\left\|T_{m}\right\|_{L^p\to L^p}\|f\|_{L^p}.$$
And $(*)$ follows.
Best Answer
The inequality is not true because the LHS and RHS have different scaling behaviour ("units").
More precisely, define a dilation operator $(δ_α f)(x) = f(α^{-1} x)$. Then by change of variables $\lVert δ_α f \rVert_{L^1} = α \lVert f \rVert_{L^1}$ and $\widehat{δ_α f} = α\, δ_{α^{-1}} \hat{f}$.
Now consider replacing $m'$ by $δ_{α^{-1}} m'$ and $f$ by $δ_α f$. For the LHS we then get \begin{align*} \renewcommand{\d}{\mathop{}\!d} &\left( \int_\mathbb{R} \left| \int_{\mathbb{R}} χ_{(t, \infty)}(ξ) m'(α t) α\hat{f}(α ξ) \d t \right|^p \d ξ\right)^{1/p} \\ &\quad= \left( \int_\mathbb{R} \left| \int_{\mathbb{R}} χ_{(α t, \infty)}(α ξ) m'(α t) \hat{f}(α ξ) \d (α t) \right|^p α^{-1} \d (α ξ) \right)^{1/p} \\ & \quad = α^{-1/p} \cdot \text{(original LHS)} \end{align*}
Meanwhile the RHS of your inequality is unchanged by this replacement. So letting $α \to 0$ we can make the LHS arbitrarily large while keeping the RHS constant so no such $C_p$ can exist (except for $p = \infty$ I suppose).