Since the Lebesgue measure $\lambda$ on $\mathbb{R}^{d}$ is atom-less,
it has the following property: For every measurable set $A$ and any
$0<\alpha<\lambda\left(A\right)$, there is a measurable subset $B\subset A$
with $\lambda\left(B\right)=\alpha$. This can be proven for example
by considering the function
$$
f:\left[0,\infty\right)\to\left[0,\infty\right),t\mapsto\lambda\left(A\cap\left[-t,t\right]^{d}\right).
$$
Note that $f$ is continuous since for $0\leq t\leq s$, we have
\begin{eqnarray*}
\left|f\left(s\right)-f\left(t\right)\right| & = & f\left(s\right)-f\left(t\right)\\
& = & \lambda\left(\left[A\cap\left[-s,s\right]^{d}\right]\setminus\left[A\cap\left[-t,t\right]^{d}\right]\right)\\
& \leq & \lambda\left(\left[-s,s\right]^{d}\setminus\left[-t,t\right]^{d}\right)\\
& = & \left(2s\right)^{d}-\left(2t\right)^{d}.
\end{eqnarray*}
Now apply the intermediate value theorem.
Let us set $\alpha:=\lambda\left(\Omega\right)\in\left(0,\infty\right)$.
By applying the above property, we see that there are pairwise disjoint(!)
subsets $\left(\Omega_{n}\right)_{n\in\mathbb{N}}$ of $\Omega$ satisfying
$\lambda\left(\Omega_{n}\right)=\frac{\alpha}{2^{n}}$ for all $n\in\mathbb{N}$.
Now, I assume that $\varphi:\mathbb{R}\to\mathbb{R}$ has the property
that the map
$$
\Phi_{\varphi}:L^{p}\left(\Omega\right)\to L^{p}\left(\Omega\right),f\mapsto\varphi\circ f
$$
is well-defined. To prove your claim, it suffices to show two separate
claims, namely:
$\forall\beta>0\exists C_{\beta}>0\forall\left|x\right|\leq\beta:\:\left|\varphi\left(x\right)\right|\leq C_{\beta}$.
$\exists\beta>0\,\exists K>0\,\forall\left|x\right|>\beta:\:\left|\varphi\left(x\right)\right|\leq K\cdot\left|x\right|$.
In fact, if we combine both claims, we get for $\left|x\right|\leq\beta$
that $\left|\varphi\left(x\right)\right|\leq C_{\beta}\leq K\left|x\right|+C_{\beta}$
and for $\left|x\right|>\beta$ likewise $\left|\varphi\left(x\right)\right|\leq K\left|x\right|\leq K\left|x\right|+C_{\beta}$.
Now, assume that the first property is false. This yields some $\beta>0$
such that the following holds:
$$
\forall n\in\mathbb{N}\exists x_{n}\in\mathbb{R}\text{ with }\left|x_{n}\right|\leq\beta\text{ and }\left|\varphi\left(x_{n}\right)\right|\geq2^{n/p}.
$$
Now, define $f:=\sum_{n=1}^{\infty}x_{n}\cdot1_{\Omega_{n}}$. Since
the $\left(\Omega_{n}\right)_{n\in\mathbb{N}}$ are pairwise disjoint(!),
we have
$$
\left\Vert f\right\Vert _{L^{p}}^{p}=\sum_{n=1}^{\infty}\left|x_{n}\right|^{p}\lambda\left(\Omega_{n}\right)\leq\beta^{p}\alpha\cdot\sum_{n=1}^{\infty}2^{-n}=\beta^{p}\alpha<\infty
$$
and hence $f\in L^{p}\left(\Omega\right)$. But we also have (using
again the disjointness) $\varphi\circ f=\sum_{n=1}^{\infty}\varphi\left(x_{n}\right)1_{\Omega_{n}}$
and thus
$$
\infty>\left\Vert \Phi_{\varphi}\left(f\right)\right\Vert _{L^{p}}^{p}=\left\Vert \varphi\circ f\right\Vert _{L^{p}}^{p}=\sum_{n=1}^{\infty}\left|\varphi\left(x_{n}\right)\right|^{p}\lambda\left(\Omega_{n}\right)\geq\sum_{n=1}^{\infty}2^{n}\frac{\alpha}{2^{n}}=\sum_{n=1}^{\infty}\alpha=\infty,
$$
a contradiction.
Hence, we have established the first property.
Now, we will show that the second property also holds, with $\beta=2^{1/p}$.
Assume that this is false. Hence, taking $K=n^{2/p}$ for arbitrary
$n\in\mathbb{N}$, we get
$$
\forall n\in\mathbb{N}\exists x_{n}\in\mathbb{R}\text{ with }\left|x_{n}\right|>2^{1/p}\text{ and }\left|\varphi\left(x_{n}\right)\right|\geq n^{2/p}\left|x_{n}\right|\geq2^{1/p}n^{2/p}.
$$
In particular, we get $\left|\varphi\left(x_{n}\right)\right|\xrightarrow[n\to\infty]{}\infty$.
In view of property 1 from above, this easily entails $\left|x_{n}\right|\xrightarrow[n\to\infty]{}\infty$,
since property 1 means that $\varphi$ is bounded on bounded sets.
Now, define
$$
N_{n}:=\left\lceil \log_{2}\left(\left|x_{n}\right|^{p}\cdot n^{2}\right)\right\rceil \xrightarrow[n\to\infty]{}\infty.
$$
Since the sequence diverges to $\infty$, we can extract a strictly
increasing subsequence $\left(N_{n_{\ell}}\right)_{\ell\in\mathbb{N}}$.
Because this sequence is strictly increasing, the sequence $\left(\Omega_{N_{n_{\ell}}}\right)_{\ell\in\mathbb{N}}$
is pairwise disjoint. Now, define
$$
f:=\sum_{\ell=1}^{\infty}x_{n_{\ell}}1_{\Omega_{N_{n_{\ell}}}}
$$
and note as above
\begin{eqnarray*}
\left\Vert f\right\Vert _{L^{p}}^{p} & = & \sum_{\ell=1}^{\infty}\left|x_{n_{\ell}}\right|^{p}\lambda\left(\Omega_{N_{n_{\ell}}}\right)\\
& = & \sum_{\ell=1}^{\infty}\left|x_{n_{\ell}}\right|^{p}\cdot\frac{\alpha}{2^{N_{n_{\ell}}}}\\
\left(\text{since }N_{n}\geq\log_{2}\left(\left|x_{n}\right|^{p}\cdot n^{2}\right)\right) & \leq & \sum_{\ell=1}^{\infty}\left|x_{n_{\ell}}\right|^{p}\cdot\frac{\alpha}{2^{\log_{2}\left(\left|x_{n_{\ell}}\right|^{p}\cdot n_{\ell}^{2}\right)}}\\
& = & \alpha\cdot\sum_{\ell=1}^{\infty}\frac{1}{n_{\ell}^{2}}\\
& \leq & \alpha\sum_{n=1}^{\infty}\frac{1}{n^{2}}<\infty.
\end{eqnarray*}
Now, note that
$$
N_{n}\leq\log_{2}\left(\left|x_{n}\right|^{p}\cdot n^{2}\right)+1
$$
and thus
$$
2^{N_{n}}\leq2\cdot\left|x_{n}\right|^{p}\cdot n^{2}.
$$
Thus, we get
\begin{eqnarray*}
\left\Vert \varphi\circ f\right\Vert _{L^{p}}^{p} & = & \sum_{\ell=1}^{\infty}\left[\left|\varphi\left(x_{n_{\ell}}\right)\right|^{p}\cdot\lambda\left(\Omega_{N_{n_{\ell}}}\right)\right]\\
& \geq & \sum_{\ell=1}^{\infty}\left[n_{\ell}^{2/p}\left|x_{n_{\ell}}\right|\right]^{p}\frac{\alpha}{2^{N_{n_{\ell}}}}\\
& \geq & \sum_{\ell=1}^{\infty}\left[n_{\ell}^{2/p}\left|x_{n_{\ell}}\right|\right]^{p}\frac{\alpha}{2\cdot\left|x_{n_{\ell}}\right|^{p}\cdot n_{\ell}^{2}}\\
& = & \frac{\alpha}{2}\cdot\sum_{\ell=1}^{\infty}1=\infty,
\end{eqnarray*}
in contradiction to $\Phi_{\varphi}\left(f\right)=\varphi\circ f\in L^{p}\left(\Omega\right)$.
Thus, we have also established property 2 from above.
As explained above, this completes the proof.
