[UPDATED IN VIEW OF COMMENTS] The answer is yes if $\mu$ is ergodic, but no in general.
For the positive result, it will suffice to show that for any continuous $g$, that for ${\mathbb P}$-almost all $x$, one has $\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(x)) \to \int g\ d\mu$. (However, the argument does not work with each $g$ separately; we will need to apply the hypothesis for other continuous observables than the original choice $g$.)
The measures $\frac{1}{k_n} \sum_{i=0}^{k_n-1} f_*^i \nu$ are increasingly $f$-invariant by telescoping series, so $\mathbb{P}$ is $f$-invariant. In particular, by the pointwise ergodic theorem, $\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(x))$ already converges pointwise $\mathbb{P}$-a.e. to some bounded measurable limit. So by dominated convergence, it will suffice to show convergence in mean, i.e.,
$$ \int |\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(x)) - \int g\ d\mu|\ d{\mathbb P}(x) \to 0$$
as $N \to \infty$.
If one chooses $\varepsilon>0$, then $N$ sufficiently large depending on $\varepsilon$, then $n$ sufficiently large depending on $N,\varepsilon$, then by weak convergence
$$ \int |\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(x)) - \int g\ d\mu|\ d{\mathbb P}(x) \leq
\int |\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(x)) - \int g\ d\mu|\ d\frac{1}{k_n} \sum_{j=0}^{k_n-1} f_*^j \nu(x) + \varepsilon
$$
$$ = \int \frac{1}{k_n} \sum_{i=0}^{k_n-1} |\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(f^j x)) - \int g\ d\mu|\ d\nu(x) + \varepsilon.$$
For $\nu$-almost every $x$, and fixed $N$, the average $\frac{1}{k_n} \sum_{i=0}^{k_n-1} |\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(f^j x)) - \int g\ d\mu|$ converges to $\int |\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(y))-\int g\ d\mu|\ d\mu(y)$ as $n \to \infty$ by hypothesis (here we are using the continuous observable $y \mapsto |\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(y))-\int g\ d\mu|$). Hence by dominated convergence, the above expression is at most
$$ \int |\frac{1}{N} \sum_{i=0}^{N-1} g(f^i(y))-\int g\ d\mu|\ d\mu(y) + 2\varepsilon$$
for $n$ large enough depending on $N,\varepsilon$. By the mean ergodic theorem and the hypothesis of ergodicity, this expression is at most $3\varepsilon$ for $N$ large enough (note that $\mu$ has to be invariant, being the weak limit of almost invariant measures $\frac{1}{N} \sum_{i=0}^{N-1} \delta_{f^i x}$ for $\nu$-almost all $x$). The claim follows.
In the non-ergodic case, the same argument shows that the claim will fail. A concrete example is the Bernoulli shift $X = \{0,1\}^{\bf Z}$ with the usual shift $f$. Let $x$ be a $0$ and $1$ sequence that alternates between long blocks of 0s and 1s with equal density of each digit, e.g., $x_n$ could be $1$ when $\lfloor \sqrt{|n|} \rfloor$ is odd, and $0$ otherwise. One can check that $\nu = \delta_x$ is generic for the measure $\mu = \frac{1}{2} \delta_0 + \frac{1}{2} \delta_1$. One can also check that any weak limit ${\mathbb P}$ of averages of $f^*_i x$ is again equal to $\mu$. But $\mu$, being non-ergodic, is not generic with respect to itself: $\mu$-almost every $y$ is equal to either $0$ or $1$, and the averages of $\delta_y$ then converge to themselves, rather than to $\mu$.
