$p_i$ is the probability of the occurence of unique symbol $i$. In the case of Tent map, the Lyapunov exponent (LE) is log of the derivative of the tent map which is almost always 2. So, $\lambda = log(2)$.
For the probabilistic Bernoulli map, I tried to find the derivative like this: $f'(x) = 1/p_1$ for $0<x<p_1$, or $f'(x) = 1-p_1/p_2$ for $p_1<x\le p_2$ and so on. Please correct me if the derivative and limits are wrong.
In K Feltekh, D Fournier-Prunaret, S Belghith, Analytical expressions for power spectral density issued from one-dimensional continuous piecewise linear maps with three slopes. Signal Process. 94:, 149–157 (2014).
the anlytical expression for LE for any piecewise linear map in general is $\lambda = (1-p)\ln(2/(1-p)) + p\ln(1/p)$ where $p$ is a constant and $p \in (0,1)$.
Then, How can I apply the above result and calculate the LE for the map in Eq(6)? Since, Tent map is conjugate to Bernoulli map, so would the probabilistic Bernoulli map have the same LE as Tent Map which is log(2)?
The Lyapunov exponent is $$ \sum_{i=1}^m p_i \log (f'|I_i)=\sum_{i=1}^m p_i \log (1/p_i)=\text{entropy}. $$
More precisely, this is the averaged Lyapunov exponent. Let me explain. You want to compute $$ \int_0^1 \lambda(x)\,d\mu(x), $$ where $\mu$ is the Bernoulli measure and $$ \lambda(x)=\lim_{n\to\infty}\frac1n\log |(f^n)'(x)| $$ when the limit exists. But since $\mu$ is invariant you get $$ \begin{split} \int_0^1 \lambda(x)\,d\mu(x) &=\lim_{n\to\infty}\frac1n\int_0^1\log |(f^n)'(x)|\,d\mu(x)\\ &=\lim_{n\to\infty}\frac1n\sum_{j=0}^{n-1}\int_0^1\log |f'(f^j(x))|\,d\mu(x)\\ &=\int_0^1\log |f'(x)|\,d\mu(x)\\ &=\sum_{i=1}^m p_i \log (f'|I_i). \end{split} $$ Indeed, since $\mu$ is $f$-invariant we have $$ \int_0^1\log |f'(f^j(x))|\,d\mu(x) =\int_0^1\log |f'(x)|\,d\mu(x) $$ for any integer $j\ge0$.