Let {$X_1, X_2,...,X_n$} be a random sample from a population with the following pmf
$$f(x|\theta,p)= \left\{ \begin{array}{lcc} (1-p)p^{x-\theta} & x=\theta, \theta+1,... \\ \\ 0 &\mbox{ otherwise}, \\ \\ \end{array} \right.$$ where $\theta$ is an positive integer and $0<p<1$, both are unknown. Find a sufficient statistic for the parameter vector $(\theta,p)$.
My Approach
Using Neyman-Fisher Factorization $$ f_X(x|\theta, p) = \prod_{i=1}^n (1-p)p^{x_i-\theta} = (1-p)^{n} p^{\sum_{i=1}^n x_i-n\theta} I_{(x_i \in {\theta,\theta+1,... })}. $$ This can be written $$ f_X(x|\theta,p) = g(T_1(x),T_2(x);\theta,p)\, h(x) \, , $$ where $g(T_1(x),T_2(x))=p^{\sum_{i=1}(x_i-n\theta)}$ and $h(x) = (1-p)^n$, which demonstrates that $(X_{(1)},\sum_{i=1}^n X_i)$ is a sufficient statistic for the parameter $(\theta,p)$.
Is the above correct? What am I missing?
Your result is correct, but the treatment of the indicators in the working is wrong, and you have made things messier than they need to be. It would be much simpler to write the likelihood function in terms of the sample mean $\bar{x} = \sum_i x_i /n$ and the sample minimum $x_{(1)}$. For all $\mathbf{x} \in \mathbb{N}^n$ we can write the likelihood function as:
$$L_\mathbf{x}(p, \theta) = (1-p)^n \cdot p^{n(\bar{x} - \theta)} \cdot \mathbb{I}(x_{(1)} \geqslant \theta).$$
Every term in this factorisation contains at least one of the parameters $p$ and $\theta$ so we cannot remove any of the terms to a function $h(\mathbf{x})$. Hence, the sufficient statistic for $(p, \theta)$ is $(\bar{x}, x_{(1)})$. If you prefer you can use the sample total to say equivalently that $(n \bar{x}, x_{(1)})$ is sufficient.