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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(\mathbb Z)\subset\ell^{2}(\mathbb Z). $