It is still discrete though not uniformly. Since $\log|P|=\frac 12\log|p|$ where $p=P\bar P$ is a real non-negative trigonometric polynomial with integer coefficients, it is enough to work with $p$ instead of $p$. We have $$ |\log p-\log q|\ge \frac{|p-q|}{\max(p,q)} $$ Now consider the outer function $f$ with $|f|=\frac 1{\max(p,q)}$, so $f(0)\ge \exp(-\int_{\mathbb T}[|\log p|+|\log q|])$. Then, denoting by $r$ the difference $p-q$ multiplied by an appropriate power of $z$ so that $r(0)\in\mathbb Z\setminus\{0\}$ and $r$ is analytic, we get $$ \int_{\mathbb T}|\log p-\log q|\ge \int_{\mathbb T}\frac{|p-q|}{p+q} =\int_{\mathbb T}|r||f|\ge |r(0)||f(0)|\ge \exp(-\int_{\mathbb T}[|\log p|+|\log q|])\,, $$finishing the story.

It is still discrete though not uniformly. Since $\log|P|=\frac 12\log|p|$ where $p=P\bar P$ is a real non-negative trigonometric polynomial with integer coefficients, it is enough to work with $p$ instead of $P$. We have $$ |\log p-\log q|\ge \frac{|p-q|}{\max(p,q)}. $$ Now consider the outer function $f$ with $|f|=\frac 1{\max(p,q)}$, so $$ |f(0)| = \exp\Big(-\int_{\mathbb T}\log{\max(p,q)}\Big) \ge \exp\Big(-\int_{\mathbb T}[|\log p|+|\log q|]\Big). $$ Then, denoting by $r$ the difference $p-q$ multiplied by an appropriate power of $z$ so that $r(0)\in\mathbb Z\setminus\{0\}$ and $r$ is analytic, we get $$ \int_{\mathbb T}|\log p-\log q|\ge \int_{\mathbb T}\frac{|p-q|}{\max(p,q)} =\int_{\mathbb T}|r||f|\ge |r(0)||f(0)| \\ \ge \exp\Big(-\int_{\mathbb T}[|\log p|+|\log q|]\Big)\,, $$ finishing the story.