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 $\mathbb{Q}$.

bounded measurablefunctions on $\Omega$ (not continuous functions; $\Omega$ has not been given a topology.) For Mike Jury's comment, the space of all probability measures on $[0,1]$ is not compact in this topology, and for Gerald Edgar's, $x \mapsto \delta_x$ is not continuous. – Nate Eldredge May 19 '13 at 13:28