Let $Z$ and $Y$ be two 0-1 decision variables.
Is there any way to approximate the following term using Taylor approximation or any other approximations?
$$\sum_w \sum_s \sum_b \left(Z_{wbr}^s \sum_{w' \neq w} p(w') Y_{w'br'} \right)$$
where $r,r'=\{1,2\}$, with a condition if $r=1$ in $Z$, $r'$ should be equal $2$ in $Y$ and vice versa. i.e. $r$ and $r'$ are not equal.
Range of $w$ is $1,\dots,34$. Range of $b$ is $1,\dots,158$. Range of $s$ is $1,2,3$.
$p(w')$ is the probability of $w'$ with values between $0$ and $0.05$
The constraint is $\displaystyle \sum_w Y_{wbr}=1, \forall b, \forall r$
Introducing a new variable with related constraints is not helpful as the number of variables of the problem will be high and the problem becomes so complicated.
You can linearize the product of two binary variables exactly by introducing one more variable and three constraints. See this Operations Research StackExchange post.
Alternatively, let $S_w = \sum_{w' \neq w} Y_{w', w}$ and linearize the product of $Z_w$ and $S_w$, as formulated here.
Updated answer based on updated question:
Introduce decision variable $X_{w,b,r,r'}^s$ to represent $Z_{wbr}^s \sum_{w' \neq w} p(w') Y_{w'br'}$, so that $$\sum_w \sum_s \sum_b \left(Z_{wbr}^s \sum_{w' \neq w} p(w') Y_{w'br'} \right) = \sum_w \sum_s \sum_b X_{w,b,r,r'}^s.$$ We want to enforce $$ X_{w,b,r,r'}^s = \begin{cases} \displaystyle{\sum_{w' \neq w} p(w') Y_{w'br'}} &\text{if $Z_{wbr}^s=1$}\\ 0 &\text{otherwise} \end{cases} $$ The following linear constraints do the job: $$ 0 \le \sum_{w' \neq w} p(w') Y_{w'br'} - X_{w,b,r,r'}^s \le 0.05(1-Z_{wbr}^s)\\ 0 \le X_{w,b,r,r'}^s \le 0.05 Z_{wbr}^s $$