Ok, Loïc Teyssier, Gerry Myerson, Nik Weaver, Stefan Waldmann, coudy, and a few other people will scold me badly for answering a question about a relatively simple exercise in undergraduate probability instead of crying loudly that "We do research here!" and closing, but here goes.
Note first of all that all examiners have an irritating habit of write the conditions in a slick form that hides what is going on as much as possible. So your first task on any exam is to trade beauty for clarity.
Since $2\sqrt(xy)=x+y-(\sqrt x-\sqrt y)^2$ and $(\sqrt x-\sqrt y)^2\asymp\frac{(x-y)^2}{x+y}$, the condition really is
$$
\sum_k (p_k-q_k)^2\left[\frac 1{(p_k+q_k)}+\frac{1}{(1-p_k)+(1-q_k)}\right]<+\infty
$$
Let $\Delta_k=q_k-p_k$.
Now the formal density of $Q$ with respect to $P$ is
$$
\prod_k \left(1+\frac{\Delta_k}{p_k}\right)^{\omega_k}\left(1-\frac{\Delta_k}{1-p_k}\right)^{1-\omega_k}
$$
We want to show that the partial products $D_n=\prod_{k=1}^n(\dots)$ converge in $L^1(P)$. The trivial sufficient condition would be the uniform $L^2(P)$ bound and we almost have it. Indeed,
$$
\mathcal E_P D_n^2=\prod_{k=1}^n\left[1+\Delta_k^2\left(\frac 1{p_k}+\frac{1}{1-p_k}\right)\right]
$$
So everything would work if we had the series without $q$'s in the denominator convergent. Unfortunately, we do not. Note, however, that if $\frac{\Delta_k^2}{p_k}>10\frac{\Delta_k^2}{p_k+q_k}$, then the quantity in the right is at least $\frac 12q_k$, so the sum over such $k$ of $q_k$ converges and, thereby, with $Q$-probability $1$ only finitely many of such $\omega_k$ are $1$. Similarly for $1-$ fractions. Thus, if we just condition upon the corresponding finite sets of bad indices (and finite subsets of integers are countably many, so we just partition the interesting part of the probability space into countably many chunks), we can use the $L^2$ criterion in each chunk separately.
This is sufficiency. Now, since it is an undergraduate exercise, do the other part yourself (or ask the people mentioned above to help you). I'll stop here.
