$\newcommand{\R}{\mathbb{R}}$
Let $Y:=X^{truncated}$ and $Z:=\Lambda$. Then, by rescaling, without loss of generality $Z\sim N(0,1)$. 

We have to show that $P(Y+Z>a)\le P(X+Z>a)$ for all $a>0$. 
Here we need to assume that $Z$ is independent of $X$ and $Y$ (I gather this assumption is missing in your question, and without this assumption the desired conclusion will not hold in general). 
In view of the symmetry of the distribution of $X$, 
\begin{align*}
	2P(X+Z>a)&=2\int_\R P(X\in dx)P(Z>a-x) \\ 
&	=\int_\R P(X\in dx)P(Z>a-x)+\int_\R P(-X\in dx)P(Z>a+x) \\ 
&=	\int_\R P(X\in dx)g(x)=Eg(X),  
\end{align*}
where 
\begin{equation*}
	g(x):=g_a(x):=P(Z>a-x)+P(Z>a+x). 
\end{equation*}
Similarly, $2P(Y+Z>a)=Eg(Y)$. Therefore and because the function $g$ is even, it suffices to show that $Eg(|Y|)\le Eg(|X|)$. 

Since 
\begin{equation*}
	Eg(|Y|)=\frac{Eg(|X|)1_{|X|\le b}}{E1_{|X|\le b}}, 
\end{equation*}
the problem reduces to the inequality 
\begin{equation*}
	Eg(|X|)h(|X|)\le Eg(|X|)\,Eh(|X|), \tag{1}
\end{equation*}
where $h(x):=1_{|x|\le b}$. 

Note that the function $g$ is increasing on $[0,\infty)$, because 
\begin{equation*}
	g'(x)=\frac1{\sqrt{2\pi}}\,(e^{-(x-a)^2/2}-e^{-(x+a)^2/2})\ge0
\end{equation*}
if $a,x\ge0$. 
Also, it is obvious that the function $h$ is decreasing on $[0,\infty)$. 

Thus, (1) follows by the integral Chebyshev inequality (see e.g. [inequality (1.1)][1]), which completes the proof. 

[1]: https://www.jstor.org/stable/4355490?seq=1#page_scan_tab_contents