The problem under consideration is: when is a Markov endomorphism (one-sided shift) $T=T_P$ with transition matrix $P$, isomorphic to a Bernoulli endomorphism $\tilde{T}_\rho$ with an appropriate stationary vector $\rho=\{\rho_i\}_{i\in I}$? An obvious necessary condition is that there exists an independent complement $\delta$ of the measurable partition $T^{-1}\varepsilon$ with $\distr \delta = \rho$. In this case the cofiltration (decreasing sequence of measurable partitions) $\xi(T)=\{T^{-n}\varepsilon\}^{\infty}_{n=1}$ generated by $T$ is finitely isomorphic to the standard Bernoulli cofiltration $\xi(\tilde{T}_{\rho})=\{\tilde{T}^{-n}_{\rho}\varepsilon\}^{\infty}_{n=1}$ and $T$ is called finitely $\rho$-Bernoulli.

We show that every ergodic Markov endomorphism $T_P$, which is finitely Bernoulli, can be represented as a skew product over $\tilde{T}_{\rho}$ with $d$-point fibres $(d\in \mathbb{N})$. We compute the minimal $d=d(T_P)$ in these skew-product representations by means of the transition matrix, and obtain necessary and sufficient conditions under which $d(T_P)=1$, i.e. $T_P$ is isomorphic to $\tilde{T}_{\rho}$. The cofiltration $\xi(T)$ of any finitely Bernoulli ergodic Markov endomorphism $T=T_P$ is represented as a $d$-point extension of the standard cofiltration $\xi(\tilde{T}_{\rho})$, and we show that the minimal $d=d_{\xi}(T)$ in these extensions is equal to $d(T)$. In particular, $d(T)=1\Longleftrightarrow d_{\xi}(T)=1$, that is, a Markov endomorphism $T$ is isomorphic to $\tilde{T}_{\rho}$ iff $\xi(T)$ is isomorphic to the Bernoulli cofiltration $\xi(\tilde{T}_{\rho})$.