The following counterexamples are taken from examples 6.1.10 and 6.1.11 in the textbook L. Grafakos, Classical Fourier Analysis (Third Edition).
Claim 1. Fix $1<p<\infty$ and $q<2$. Then the inequality
$$\| (\sum_{j\in\mathbb{Z}} |P_{j}(f)|^{q})^{1/q}\|_{L^{p}} \lesssim_{p,q} \|f\|_{L^{p}}$$
cannot hold.
Proof. Let $\psi$ be a $C^{\infty}$ function on $\mathbb{R}$ such $\hat{\psi}\geq 0$,
$$\hat{\psi}(\xi) = \begin{cases} 1 & {\xi \in [9/8,15/8]} \\ 0 & {\xi\notin [7/8,17/8]} \end{cases},$$
and
$$\sum_{j\in\mathbb{Z}} \hat{\psi}(2^{-j}\xi)^{2}=1, \qquad \xi>0.$$
Extend $\hat{\psi}$ to be an even function on $\mathbb{R}$. Let $P_{j}:=\psi(2^{-j}D)$.
Now let $\varphi$ be a Schwartz function such that $\hat{\varphi}\geq 0$ and $\mathrm{supp}(\hat{\varphi}) \subset [11/8,13/8]$. Fix $N\gg 1$, and define
$$f_{j}(x) := e^{2\pi\frac{12}{8}2^{j}x}\varphi(x), \qquad j=1,\ldots,N.$$
Since $\widehat{f}_{j}(\xi) = \hat{\varphi}(\xi-\frac{12}{8}2^{j})$ is supported on the set
$$[\frac{11}{8} + \frac{12}{8}2^{j}, \frac{13}{8}+\frac{12}{8}2^{j}]\subset [\frac{9}{8}2^{j}, \frac{15}{8}2^{j}], j\gg 1,$$
$\hat{\psi}(2^{-j}\xi)=1$ on the support of $\hat{f}_{j}$.
Now define
$$f := \sum_{j=3}^{N}f_{j}.$$
We have the lower bound
$$\|(\sum_{j}|P_{j}(f)|^{q})^{1/q}\|_{L^{p}}\geq \|(\sum_{j=3}^{N} |f_{j}|^{q})^{1/q}\|_{L^{p}} = (N-3)^{1/q}\|\varphi\|_{L^{p}}.$$
By the Littlewood-Paley inequality, we have the upper bound
$$\|f\|_{L^{p}} \lesssim_{p} \|(\sum_{j}|P_{j}f|^{2})^{1/2}\|_{L^{p}} = (N-3)^{1/2}\|\varphi\|_{L^{p}}.$$
For $N\gg 1$, we obtain a contradiction. $\Box$
Claim 2. Fix $1<p<\infty$ and $2<q<\infty$. Then the inequality
$$\|g\|_{L^{p}} \lesssim_{p,q} \|(\sum_{j} |P_{j}g|^{q})^{1/q}\|_{L^{p}}$$
cannot hold.
Proof. This follows from duality and Claim 1. $\Box$.