Bernoulli Shift has complete positive entropy without using K-automorphism property

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Let $X = (1,\dots,d)^{\mathbb{Z}}$ with the $\sigma$-álgebra $\beta $ generated by the cylinders and $\mu$ the product measure.

Let $T: X \rightarrow X$ be the left shifter, $T(x(n))=x(n+1)$.

We define the entropy of a partition $\mathcal{H}(\mathcal{P})=\displaystyle\sum_{P \in \mathcal{P}}-\mu(P)\ln(\mu(P))$ and the entropy of $T$ relataded to $\mathcal{P} $, $ h(T,\mathcal{P})= \displaystyle\lim_{n \rightarrow \infty} \frac{1}{n}\mathcal{H}(\bigvee_{i=0}^{n-1}T^{-i}\mathcal{P})$.

I am able to show that $T$ is a $K$-automorphism, and i want to prove that $T$ has complete positive entropy(CPE), i.e. ,if $\mathcal{P}$ is a non trivial enumerable partition of $X$ with finite entropy, so $h(T,\mathcal{P}) > 0$.

Since shift have both $K$-automorphism property and CPE, it's interesting to see if $K$-automorphism property and CPE are equivalent, and that holds at last when $(X,\beta,\mu)$ is a separable probability space.

Again i am able to show the equivalence using the bibliograph: Topics in ergodic theory by W.Parry, but since what motivates the equivalence is that the shifts have both proprety I dont want to use the equivalence to show that shift have CPE.

A hint that was give to me is to show first that if $\mathcal{P}$ is a finite partition of $X$ so $\displaystyle\bigcap_{n=1}^{\infty}T^{-n}(\bigvee_{i=1}^{\infty}T^{-i}\widehat{\mathcal{P}}) =\{\varnothing, X\}$, where $\widehat{\mathcal{P}} = \sigma(\mathcal{P})$, and use that to show the CPE property.

I use too the bibliograph:Foundations of Ergodic Theory by Marcelo Viana and Krerley Oliveira.

Any help will be appreciated.