How can I start thinking about this?
Let $X$, $Y$ be i.i.d. random variables. Suppose $θ$ is independent of ($X$, $Y$), with $P(θ = 1) = p$, $P(θ = 0) = 1 − p$. Let $Z = (Z_1,Z_2)$ where $Z_1 = θX + (1 − θ)Y$ and $Z_2 = θY + (1 − θ)X$.
Obtain an explicit expression for $E[g(X, Y )|Z]$, in terms of $Z_1$ and $Z_2$, where $g : R^2 \rightarrow R$ is a bounded Borel function.
Since $\theta$ takes the values $0$ and $1$ only, then $\theta(1-\theta)=0$, $\theta^2=\theta$, $(1-\theta)^2=1-\theta$, so you can multiply $Z_1$ and $Z_2$ by $\theta$ and $(1-\theta)$ and get four equalities: $$\theta Z_1 = \theta X, \tag{1}$$ $$(1-\theta)Z_1 = (1-\theta) Y, \tag{2}$$ $$(1-\theta)Z_2 = (1-\theta) X, \tag{3}$$ $$\theta Z_2 = \theta Y.\tag{4}$$ Add (1) + (3), then (2)+(4): $$ \theta Z_1 + (1-\theta)Z_2 = \theta X + (1-\theta) X = X, $$ $$ \theta Z_2 + (1-\theta)Z_1 = \theta Y + (1-\theta) Y = Y. $$ So $g(X,Y)=g(\theta Z_1 + (1-\theta)Z_2,\theta Z_2 + (1-\theta)Z_1)$. And you need to find $$ \mathbb E\bigl[g(\theta Z_1 + (1-\theta)Z_2,\,\theta Z_2 + (1-\theta)Z_1) \bigm | Z_1,Z_2\bigr]. $$ Since $\theta$ takes only values $0$ and $1$, the answer follows immediately.