I think it is not $G_\delta$.
Take $X = [0,1]$ for instance and let $m$ denote Lebesgue measure. First, note that $\mathcal{P}_0$ is dense in $\mathcal{P}$.
Now I claim it is possible to find a homeomorphism $F$ of $[0,1]$ and a Borel set $E \subset [0,1]$ such that $m(E) = 1$ and $m(F(E)) = 0$.
(For instance, let $\mu_0$ be an atomless probability measure of full support which is singular to Lebesgue measure, so that there exists a Borel set $E$ with $m(E) = 1$ and $\mu_0(E)=0$, and let $F(x) = \mu_0([0,x])$ be the cdf of $\mu_0$, which one can check is a homeomorphism. Then we have $m(F(A)) = \mu_0(A)$ for $A$ of the form $[0,x]$, hence for all Borel sets $A$, and taking $A=E$ we have what we want.)
Let $G = F^{-1}$ which is also a homeomorphism, and for $\mu \in \mathcal{P}$ let $\Gamma\mu = G_* \mu$ be the pushforward of $\mu$ under $G$, i.e. $(\Gamma\mu)(A) = \mu(G^{-1}(A)) = \mu(F(A))$. It is a standard exercise to check that $\Gamma$ is a homeomorphism of $\mathcal{P}$. Moreover, if $\mu \in \mathcal{P}_0$ then we have $\mu(E) = 1$ but $\Gamma \mu(E) = \mu(F(E)) = 0$, so that $\Gamma \mu$ is singular to $m$. In particular, $\Gamma \mathcal{P}_0 \cap \mathcal{P}_0 = \emptyset$. So if $\mathcal{P}_0$ were $G_\delta$, then $\mathcal{P}_0$ and $\Gamma \mathcal{P}_0$ would be two disjoint dense $G_\delta$ sets in $\mathcal{P}$, which contradicts the Baire category theorem.
Actually, we can soup this up to prove that the set is meager. On $[0,1]$, we may find an uncountable family $\{\mu_i : i \in I\}$ of probability measures which are atomless, full support, and mutually singular, and you may if you like take one of them to be Lebesgue measure. A standard example is to take $I = (0,1)$ and let $\mu_i$ be the law of a random variable $X_i = \sum_n 2^{-n} \xi_{i,n}$ where the $\xi_{i,n}, n = 1, 2, \dots$ are iid Bernoulli($i$). And by considering cdfs as above, for each $i,j$ there is a homeomorphism $F_{ij}$ of $[0,1]$ such that $\mu_j = \mu_i \circ F_{ij}^{-1}$. Hence the pushforward maps $\Phi_{ij} := (F_{ij})_*$ are homeomorphisms of $\mathcal{P}$.
Let $\mathcal{P}_i$ be the set of all probability measures absolutely continuous to $\mu_i$. Note the sets $\mathcal{P}_i$ are disjoint. I claim $\Phi_{ij} \mathcal{P}_i = \mathcal{P}_j$. For suppose $\nu_i \ll \mu_i$ and $\mu_j(E) = 0$. Then $\mu_i(F_{ij}^{-1}(E)) = \mu_j(E) = 0$ so we have $\Phi_{ij} \nu_i(E) = \nu_i(F_{ij}^{-1}(E)) = 0$, which is to say that $\Phi_{ij} \nu_i \ll \mu_j$. The reverse inclusion follows by symmetry.
So suppose one of the sets $\mathcal{P}_i$ is nonmeager. Then all of them are nonmeager. However, they are all Borel as you noted, and therefore they have the property of Baire: each $\mathcal{P}_i$ can be written as $\mathcal{P}_i = \mathcal{U}_i \triangle \mathcal{M}_i$ where $\mathcal{U}_i$ is open and $\mathcal{M}_i$ is meager. By assumption, each $\mathcal{U}_i$ must be nonempty. Now for $i \ne j$, since $\mathcal{P}_i \cap \mathcal{P}_j = \emptyset$, we have $\mathcal{U}_i \cap \mathcal{U}_j \subset \mathcal{M}_i \cup \mathcal{M}_j$. So $\mathcal{U}_i \cap \mathcal{U}_j$ is a meager open set, which by the Baire category theorem must be empty. Thus $\{\mathcal{U}_i\}$ is an uncountable collection of nonempty disjoint open sets, which is absurd since $\mathcal{P}$ is separable. We conclude, then, that all the sets $\mathcal{P}_i$ are meager.
Take $A_n$ sequence of increasing sets of $\mathcal{A}$ such that $A = \underset{n \geq
1}{\bigcup}A_n \in \mathcal{A}$. We wish to show that $\lim \mu(A_n) = \mu(A)$
Since $A_n \subset A, \forall n \geq 1$, then $\mu(A_n) \leq \mu(A) \implies \lim \mu(A_n) \leq \mu(A)$. For the other inequality, we may view this increasing sequence of sets as an increasing disjoint union of sets by defining $B_1 = A_1$ and $B_n = A_n-A_{n-1}, \ n \geq 2$. We have $B_n \in \mathcal{A}, \ \forall n \geq 1$ since $\mathcal{A}$ is an algebra and $A_m = \sum\limits_{n=1}^mB_n$ where the sum notation is used to denote a union of pairwise disjoint sets. Notice that since $\mu$ is finitely additive, then $\mu(A_m) = \sum\limits_{n=1}^m\mu(B_n)$ and now we will use that the measure is regular:
Let $\epsilon>0$ and $U_n$ be an open set such that $B_n \subset U_n$ and $\mu(B_n) \leq \mu(U_n) \leq \mu(B_n) + \frac{\epsilon}{2^n}$ and let $K \subset A$ be a compact set. Notice that $K \subset A = \sum\limits_{n\geq 1}B_n \subset \underset{n\geq 1}{\bigcup}U_n$, therefore by compactness of $K$, there exists $m'$ such that $K \subset \bigcup\limits_{n=1}^{m'}U_n$ which implies $\mu(K) \leq \mu(\bigcup\limits_{n=1}^{m'}U_n) \leq \sum\limits_{n=1}^{m'}\mu(U_n) \leq \sum\limits_{n\geq 1}\mu(B_n) + \epsilon$, since $\epsilon$ is arbitrary, then $\mu(K) \leq \sum\limits_{n\geq 1}\mu(B_n) = \underset{m \to \infty}{\lim}\sum\limits_{n=1}^m\mu(B_n) = \underset{m \to \infty}{\lim}\mu(A_m)$, since this is true for every compact set contained in $A$, and $\mu(A)$ is the supremum of the measures of all such compact sets, then $\mu(A) \leq \lim\mu(A_n)$.
Best Answer
You are making things too complicated. If $P_n \to P$ in TV then $P_n(B) \to P(B)$ for every Borel set $B$. Hence if $B$ is a Borel set with Lebesgue measure $0$ then $P(B) =\lim P_n(B)=\lim 0=0$. Hence $P$ is also absolutely continuous.