If that inequality was true, for all $f\in L^p$, that would imply
$$
L^p\subset \dot B^0_{p,1}\quad\text{with continuous injection}, 
$$
with $\dot B^0_{p,1}$ the homogeneous Besov space whose norm is precisely given by the left-hand-side of your inequality. On the other hand, the reverse inequality
$$
\Vert{f}\Vert_{L^p}=\Vert{\sum_{m\in \mathbb Z}P_m f}\Vert_{L^p}\le
\sum_{m\in \mathbb Z}\Vert{P_m f }\Vert_{L^p}
$$
is true for all $f\in \dot B^0_{p,1}\cap L^p$. We would have the topological equality
$L^p= \dot B^0_{p,1}$. But it is classical
that for $1<p<\infty$,
$$
L^p=F^0_{p,2}
$$
where $F^0_{p,2}$ is a Triebel-Lizorkin space. Also classical is the fact that a Triebel-Lizorkin space is never a Besov space, except for $p=2$. Even in the case $p=2$ that inequality is false since it would imply
$$
\dot B^0_{2,2}=L^2=\dot B^0_{2,1},
$$
which is incompatible with the strict inclusion
$
\ell^1\subset\ell^{2}.
$