For linear maps like $\mu \mapsto \int_X \varphi d\mu$ where $\varphi$ is any measurable and Borel function on $X$, the conclusion holds, since for every $n$, the measure $mu_n$ equals $\lfloor np \rfloor /n$ times $\mu_p$ plus a measure with total mass at most $(p-1)/n$.

The notations are contradictory. Once $p$ is fixed, and then it varies.

Do you set $\mu_{n}:=\frac{1}{n}\sum_{i=0}^{n-1}T_{\ast}^{i}\nu$ for ALL $n \ge 1$ and assume that $T_{\ast}^{p} \nu = \nu$ for SOME fixed $p$ ?

If yes, $\mu_{p}$ is $T$-invariant and for every $n \ge 1$, calling $q_n$ and $r_n$ the quotient and the remainder of the Euclidean division by $p$, we get by $p$-periodicity of $(T_{\ast}^{i}\nu)_{i \ge 0}$ $$\sum_{i=0}^{n-1} T_{\ast}^{i}\nu = \sum_{i=0}^{pq_n-1} T_{\ast}^{i}\nu + \sum_{i=pq_n}^{n-1} T_{\ast}^{i}\nu = q_n\sum_{i=0}^{p-1} T_{\ast}^{i}\nu + \sum_{i=0}^{r_n-1} T_{\ast}^{i}\nu.$$ Dividing by $n$ we get $$\mu_n = \frac{pq_n}{n} \mu_p + \frac{r_n}{n} \mu_{r_n},$$ with the convention $\mu_0=\nu$ (actually, the choice of $\mu_0$ can be arbitrary and plays no role). If $f : M(X,T) \to \mathbb{R}$ is affine, we get $$f(\mu_n) = \frac{pq_n}{n} f(\mu_p) + \frac{r_n}{n} f(\mu_{r_n}).$$ The sequence $(f(\mu_{r_n}))_{n \ge 1}$ is periodic hence bounded, so $$f(\mu_n) \to f(\mu_p) \text{ as } n \to +\infty.$$ No other property of $f$ is required.