Question on the update step of Bayesian filter
I would prefer to see the definitions, but based on your prior work (which I haven't checked thoroughly), we can already reach the final stage. You left off at $$ p(x_k | y_{1:k}) = \frac{p(y_k|x_k) p(x_k|y_{1:k-1})}{p(y_k|y_{1:k-1})} $$ which is a probability distribution of $x_k$ based on the left hand side. The term in the denominator is independent of $x_k$ (if I have guessed your model correctly). Therefore, it is an unknown positive constant. That is, there exists some constant $Z_k$ such that $$ p(x_k | y_{1:k}) = \frac{1}{Z_k} p(y_k|x_k) p(x_k|y_{1:k-1})$$ where $Z_k = p(y_k|y_{1:k-1})$. 

Now let's integrate both sides.
$$ \int p(x_k | y_{1:k}) d x_k = \int \frac{1}{Z_k} p(y_k|x_k) p(x_k|y_{1:k-1}) d x_k $$
The left hand side is a probability distribution with respect to $x_k$, therefore it sums to $1$. Because $Z_k$ is independent of $x_k$, we can pull it out of the integral. Multiply both sides by $Z_k$ to obtain your answer,
$$Z_k = \int p(y_k|x_k) p(x_k | y_{1:k-1}) d x_k.$$

Is $Z_k$ a constant with respect to $x_k$?

We know that $Z_k = p(y_k|y_{1:k-1})$ in other words $Z_k$ is the probability of $y_k$ given all previous $y_{1:k-1}$ after integrating out $x_k$. Note that this is the same thing as $$Z_k = \int p(y_k|y_{1:k-1},x_k) p(x_k) d x_k$$
which (after integrating) removes all dependency on $x_k$. Therefore, as noted in the original response, $Z_k$ is constant with respect to $x_k$ and can be factored out of the integral.

This practice of marginalizing out a variable based on its prior is a common theme in Bayesian systems.