As noted in the OP, if $\mathcal P$ is countable, for example $\mathcal P=(\mu_n,n\in\Bbb N)$, then take $\nu:=\sum_{n=1}^{+\infty}2^{-n}\mu_n$.

Edit: This was posted before having the wanted definition of weak star topology

When $\mathcal P$ is not countable, it's not necessarily true: take $\Omega:=[0,1]$, $\mathcal F$ its Borel $\sigma$-algebra and $\mathcal P:=\(\delta_x,x\in[0,1]\)$. Then $\mathcal P$ is weak-$*$ star compact (as $x\mapsto \delta_x$ from $[0,1]$ to $\mathcal P$ is a homeomorphism), but there is no measure $\nu$ such that $\delta_x\ll\nu$ for all $x\in[0,1]$. Otherwise, $\nu(x)>0$ for all $x$, which is incompatible with additivity of $\nu$ and uncountability of the unit interval.

Considering the weak star topology as given in the OP, the answer seems to be yes. First, let $\mu\sim\nu$ mean that $\mu\ll\nu$ and $\nu\ll\mu$. We identify two equivalent measures. Then we work with Zorn's lemma with order $\mu\leq\nu$ iff $\mu\ll\nu$. Weak star compacteness allows us to show that each totally ordered subset of $\mathcal P$ admits a maximal element.

As noted in the OP, if $\mathcal P$ is countable, for example $\mathcal P=(\mu_n,n\in\Bbb N)$, then take $\nu:=\sum_{n=1}^{+\infty}2^{-n}\mu_n$.

Edit: This was posted before having the wanted definition of weak star topology

When $\mathcal P$ is not countable, it's not necessarily true: take $\Omega:=[0,1]$, $\mathcal F$ its Borel $\sigma$-algebra and $\mathcal P:=\(\delta_x,x\in[0,1]\)$. Then $\mathcal P$ is weak-$*$ star compact (as $x\mapsto \delta_x$ from $[0,1]$ to $\mathcal P$ is a homeomorphism), but there is no measure $\nu$ such that $\delta_x\ll\nu$ for all $x\in[0,1]$. Otherwise, $\nu(x)>0$ for all $x$, which is incompatible with additivity of $\nu$ and uncountability of the unit interval.

Considering the weak star topology as given in the OP, the answer seems to be yes. First, let $\mu\sim\nu$ mean that $\mu\ll\nu$ and $\nu\ll\mu$. We identify two equivalent measures. Then we work with Zorn's lemma with order $\mu\leq\nu$ iff $\mu\ll\nu$. Weak star compacteness allows us to show that each totally ordered subset of $\mathcal P$ admits a maximal element.

As noted in the OP, if $\mathcal P$ is countable, for example $\mathcal P=(\mu_n,n\in\Bbb N)$, then take $\nu:=\sum_{n=1}^{+\infty}2^{-n}\mu_n$.

When $\mathcal P$ is not countable, it's not necessarily true: take $\Omega:=[0,1]$, $\mathcal F$ its Borel $\sigma$-algebra and $\mathcal P:=\(\delta_x,x\in[0,1]\)$. Then $\mathcal P$ is weak-$*$ star compact (as $x\mapsto \delta_x$ from $[0,1]$ to $\mathcal P$ is a homeomorphism), but there is no measure $\nu$ such that $\delta_x\ll\nu$ for all $x\in[0,1]$. Otherwise, $\nu(x)>0$ for all $x$, which is incompatible with additivity of $\nu$ and uncountability of the unit interval.

As noted in the OP, if $\mathcal P$ is countable, for example $\mathcal P=(\mu_n,n\in\Bbb N)$, then take $\nu:=\sum_{n=1}^{+\infty}2^{-n}\mu_n$.

Edit: This was posted before having the wanted definition of weak star topology

When $\mathcal P$ is not countable, it's not necessarily true: take $\Omega:=[0,1]$, $\mathcal F$ its Borel $\sigma$-algebra and $\mathcal P:=\(\delta_x,x\in[0,1]\)$. Then $\mathcal P$ is weak-$*$ star compact (as $x\mapsto \delta_x$ from $[0,1]$ to $\mathcal P$ is a homeomorphism), but there is no measure $\nu$ such that $\delta_x\ll\nu$ for all $x\in[0,1]$. Otherwise, $\nu(x)>0$ for all $x$, which is incompatible with additivity of $\nu$ and uncountability of the unit interval.

Considering the weak star topology as given in the OP, the answer seems to be yes. First, let $\mu\sim\nu$ mean that $\mu\ll\nu$ and $\nu\ll\mu$. We identify two equivalent measures. Then we work with Zorn's lemma with order $\mu\leq\nu$ iff $\mu\ll\nu$. Weak star compacteness allows us to show that each totally ordered subset of $\mathcal P$ admits a maximal element.