My attempt:
Looks like the definition of $p_{n,t}$ implies that the Bernulli trials $\xi_j$ are independent (?) I will assume that, please confirm or refute.
$Z_n=(\frac{a}{b})^{\sum_{j=1}^n\xi_j}(\frac{1-a}{1-b})^{n-\sum_{j=1}^n\xi_j}:=g(S_n)$
with $S_n := \sum_{j=1}^n\xi_j \sim Bin(\theta,n)$ The natural filtration they are referencing is if I am not mistaken $\mathcal F_n=\sigma(\xi_1,...,\xi_n)$.
I have to prove that $Z_n$ is a martingale wrt to $\mathcal F_n$ iff $\theta=b$
$Z_n$ is a function of $\xi_1,...,\xi_n$ so it is $\mathcal F_n$-measurable, therefore it is adapted
$E|Z_n|=EZ_n=\sum_{s=1}^n g(s)\mathbb P (S_n=s)<\infty$ because is a finite sum with finite terms. ($S_n \sim Bin(\theta,n)$, so P (S_n=s)<\infty )
$$E[Z_{n+1}|\mathcal F_n]= E[(\frac{a}{b})^{\sum_{j=1}^{n+1}\xi_j}(\frac{1-a}{1-b})^{n+1-\sum_{j=1}^{n+1}\xi_j}|\mathcal F_n]$$ $$=E[(\frac{a}{b})^{\xi_{n+1}}(\frac{1-a}{1-b})^{-\xi_{n+1}}|\mathcal F_n] (\frac{a}{b})^{\sum_{j=1}^{n}\xi_j}(\frac{1-a}{1-b})^{n-\sum_{j=1}^{n}\xi_j}(\frac{1-a}{1-b})$$ $$=E[(\frac{a}{b})^{\xi_{n+1}}(\frac{1-a}{1-b})^{-\xi_{n+1}}] Z_n(\frac{1-a}{1-b})$$
where in the last steps I used that $\xi_1,...\xi_n$ are $\mathcal F_n$-measurable and the independence of $\xi_{n+1}$ from $\mathcal F_n$.
For this to be a martingale I need that
$$E[(\frac{a}{b})^{\xi_{n+1}}(\frac{1-a}{1-b})^{-\xi_{n+1}}] Z_n(\frac{1-a}{1-b}) = Z_n \quad a.s$$
$$E[(\frac{a(1-b)}{b(1-a)})^{\xi_{n+1}}] Z_n(\frac{1-a}{1-b}) = Z_n \quad a.s$$
At this point I have to options to compute this expectation, but I am not sure which one is the correct one:
option 1
$$ (\frac{a(1-b)}{b(1-a)})^{1}\mathbb P(\xi_{n+1}=1)+ 1.\mathbb P(\xi_{n+1}=0) = (\frac{1-b}{1-a}) \quad a.s$$
$$ (\frac{a(1-b)}{b(1-a)})^{1}\theta+ 1. (1-\theta) = (\frac{1-b}{1-a}) \quad a.s$$
From this I get $\theta=b$.
option 2
$ (\frac{a(1-b)}{b(1-a)})^{1}[\mathbb P(\xi_{n+1}=1|\theta=a) \mathbb P(\theta=a) + \mathbb P(\xi_{n+1}=1|\theta=b) P(\theta=b) ]+ 1.[\mathbb P(\xi_{n+1}=0|\theta=a) \mathbb P(\theta=a) \mathbb P(\xi_{n+1}=0|\theta=b) \mathbb P(\theta=b) ]= (\frac{1-b}{1-a}) \quad a.s $
$ (\frac{a(1-b)}{b(1-a)})^{1}[\mathbb a 0.5 + b 0.5 ]+ 1.[(1-a) 0.5 + (1-b) 0.5 ]= (\frac{1-b}{1-a}) \quad a.s $
where I have used that since there are tow options for $\theta$: $\mathbb P(\theta=b) = \mathbb P(\theta=a) =0.5$ This yields a=b which is a contradiction and the equation does not involve $\theta $
1) What do you think? While option 1 yield the correct answer, in option 2 I am conditioning on the values of theta as the problem suggests. Why is option 2 not correct?
2) Was it correct that $\mathcal F_n=\sigma(\xi_1,...,\xi_n)$ and that the $\xi_i$ are independent? I assumed $p_{n,t}$ was the joint pmf of $\xi_1,...\xi_n$ (but the problem doesn't say so) and notice that if that is the case, it is the product of the marginal pmf of the Bernulli