Of course, there are many ways of metrizing the weak topology on $\mathcal M(\Omega)$ by using various tools of functional analysis. However, as it has already been pointed out by Dan, the most natural way is to use the transportation metric on the space of measures. [It is much more natural than the Prokhorov metric. I don't want to go into historical details here - they can be easily found elsewhere, but I insist that the transportation metric should really be related with the names of Kantorovich (in the first place) and his collaborator Rubinshtein]. Dan gives its dual definition in terms of Lipschitz functions, however its "transport definition" is actually more appropriate here. Let me remind it.
Given two probability measures $\mu_1,\mu_2$ on $\Omega$
$$
\overline d(\mu_1,\mu_2) = inf_M \int d(x_1,x_2) dM(x_1,x_2) \;,
$$
where $d$ is the original metric on $\Omega$, and the infimum (which is in fact attained) is taken over all probability measures $M$ on $\Omega\times\Omega$ whose marginals ($\equiv$ coordinate projections) are $\mu_1$ and $\mu_2$. One should think about such measures as "transportation plans" between distributions $\mu_1$ and $\mu_2$, while the integral in the RHS of the definition is the "cost" of the plan $M$.
It is obvious that the above definition makes sense not just for probability measures, but for any two positive measures $\mu_1,\mu_2$ with the same mass. Moreover, $\overline d(\mu_1,\mu_2)$ actually depends on the difference $\mu_1-\mu_2$ only, so that one can think about it as a "weak norm"
$$
|||\mu_1-\mu_2||| = \overline d(\mu_1,\mu_2)
$$
of the signed measure $\mu_1-\mu_2$ (clearly, it is homogeneous with respect to multiplication by scalars).
Let now $\mu=\mu_1-\mu_2$ be an arbitrary signed measure, where $\mu_1,\mu_2$ are the components of its Hahn decomposition. The only reason why the definition of the weak norm does not work in this situation is that the measure $\mu$ need not to be "balanced" in the sense that the total masses $\|\mu_1\|$ and $\|\mu_2\|$ need not be the same any more. However, this can be easily repaired in the following way: extend the original space $\Omega$
to a new metric space $\Omega'$ by adding to it an "ideal point" $o$ and putting $d(\omega,o)=1$ for any $\omega\in\Omega$. Then the measure
$$
\mu'=\mu - (\|\mu_1\|-\|\mu_2\|)\delta_o \;,
$$
where $\delta_o$ is the unit mass at the point $o$, is now balanced, so that $|||\mu'|||$ is well defined. Therefore, one can extend the definition of the weak norm $|||\cdot|||$ to arbitrary signed measures $\mu$ by putting
$$
|||\mu|||=|||\mu'||| \;.
$$
It is now easy to see that the distance $|||\mu_1-\mu_2|||$, where $\mu_1,\mu_2$ are two arbitrary signed measures, metrizes the weak topology on $\mathcal M(\Omega)$.
Best Answer
There always exists a dominating measure.
First, given two finite measures $\mu,\nu$ on $(\Omega,\mathcal{F})$, the Lebesgue decomposition theorem says that there is an $A\in\mathcal{F}$ such that $1_{\Omega\setminus A}\mu$ is absolutely continuous with respect to $\nu$ and $1_A\mu,\nu$ are singular. This can also be constructed using the Radon-Nikodym derivative as $A=\lbrace d\nu/d(\mu+\nu)=0\rbrace$, and $A$ is uniquely defined up to a $\mu$-null set. We can think of $\mu(A)$ as measuring the distance $\mu$ is from being absolutely continuous wrt $\nu$, and I will write $d(\mu,\nu):=\mu(A)$.
Assuming $\mathcal{P}$ is nonempty then, using dependent choice, there exists a sequence $\mathbb{P}\_1,\mathbb{P}\_2,\ldots\in\mathcal{P}$ such that $$ d(\mathbb{P}\_n,\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1}) \ge \frac12\sup\left\lbrace d(\mathbb{P},\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1})\colon\mathbb{P}\in\mathcal{P}\right\rbrace. $$ We can show that $\mathbb{Q}:=\sum_{n=1}^\infty2^{-n}\mathbb{P}\_n$ is a dominating measure for $\mathcal{P}$.
Let me start by showing that $\alpha_n:=d(\mathbb{P}\_n,\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1})$ tends to zero. By definition, there exists $A_n\in\mathcal{F}$ such that $\mathbb{P}\_n(A_n)=\alpha_n$ and $\mathbb{P}\_m(A_n)=0$ for $m < n$. Replacing $A_n$ by $A_n\setminus\bigcup_{m > n}A_m$ if necessary, we can further suppose that $A_n$ are disjoint. Now, set $B_n=\bigcup_{m\ge n}A_m$ so that $\mathbb{P}\_m(B_n)\ge\alpha_m$ for $m\ge n$ and $B_n\downarrow\emptyset$ as $n\to\infty$. If $\alpha_n$ does not tend to zero then, by passing to a subsequence if necessary, it can be assumed that $\alpha_n\rightarrow\alpha > 0$. Now, by compactness, the sequence $\mathbb{P}\_m$ has a limit point $\mathbb{P}$ in the weak-* topology. By the definition of the topology given in the question, this means that $\mathbb{P}(B_n)$ is a limit point of $\mathbb{P}\_m(B_n)\ge\alpha_m$ as $m\to\infty$, so $\mathbb{P}(B_n)\ge\alpha$. This contradicts the fact that $\mathbb{P}(B_n)\to0$ as $n\to\infty$ (by countable additivity). So, $\alpha_n\to0$ as required.
Finally, lets show that $\mathbb{Q}$ is a dominating measure. Choosing any $\mathbb{P}\in\mathcal{P}$ set $\alpha:=d(\mathbb{P},\mathbb{Q})$. If $\alpha > 0$ then we would have $\alpha_n < \alpha/2$ for large $n$, in which case $$ \begin{align} d(\mathbb{P}\_n,\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1})&=\alpha_n < \alpha/2 =\frac12d(\mathbb{P},\mathbb{Q})\cr &\le\frac12d(\mathbb{P},\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1}). \end{align} $$ The last inequality uses the fact that $\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1}$ is absolutely continuous wrt $\mathbb{Q}$. This inequality contradicts the choice of $\mathbb{P}\_n$, so we have $\alpha=0$ and $\mathbb{P}$ is absolutely continuous wrt $\mathbb{Q}$.