Computing Denominator of Bayes theorem - independently vs conditionally Lets say that you have n independent variables (assume n is even) $x_1$ to $x_n$,
Let $X = [x_1, x_2, x_3, ...., x_n]$
We have another variable $y$ that can only take two values $0$ and $1$ such that
$P(y=0)=P(y=1)=0.5$,
Given that $P(x_i=1|y=1) = a$, $P(x_i=1|y=0)=b$ for all $i$ from $1$ to $n$
Find $P(y=1|X)$ where $X = [1,0,1,0,....,1,0]$.
If we use Baye's theorem, we get
$$
P(y=1|X) = \frac{P(X|y=1)P(y=1)}{P(X)}
$$
Now to compute the denominator $P(X)$,
Option 1 (compute independently)
should I write it as $P(X) = P(x_1=1)P(x_2=0)P(x_3=1)...P(x_{n}=0)$
To solve this, I first find $P(x_i=1) = P(x_i=1|y=1)P(y=1) + P(x_i=1|y=1)P(y=1)$
This gives us,
$P(x_i=1)=(a+b)/2$
and $P(x_i=0)=1-(a+b)/2$
So the final answer using this approach is 
$P(X) = ((a+b)/2)^{n/2} (1-(a+b)/2)^{n/2}$
OR
Option 2 (compute conditionally)
should I write it as $P(X) = P(X|y=1)P(y=1) + P(X|y=0)P(y=0)$
To solve this we have,
$P(X|y=1) = P(x_1=1|y=1)P(x_2=0|y=1)...P(x_n=0|y=1)$
$P(X|y=1) = a(1-a)a(1-a)...(1-a)$
Similarly
$P(X|y=0) = b(1-b)b(1-b)...(1-b)$
This gives us the final answer as
$P(X) = a^{n/2}(1-a)^{n/2}0.5 + b^{n/2}(1-b)^{n/2}0.5$
Answers from the above two options are clearly different, which one is correct?
 A: You're asking which of two options you should pick, but I think both of those ideas will be important here!
For convenience I'll assume $n$ is even. If it's odd you need to handle the end of the products carefully since $X$ will have more $1$s than $0$s.
First, for any $k$ we have
$$
\begin{align}
P(x_k = 0) &= P(x_k = 0 | y = 0) P(y=0) + P(x_k=0|y=1) P(y=1) = \frac {1-b} 2 + \frac {1-a} 2 = \frac{2-a-b}2
\end{align}$$
and
$$P(x_k = 1) = P(x_k = 1 | y = 0) P(y=0) + P(x_k=1|y=1) P(y=1) = \frac b 2 + \frac a 2 = \frac{a+b}2.$$
Now we'll use those to compute $P(X)$:
$$
\begin{align}
P(X) &= P(x_1=1)P(x_2=0)P(x_3=1)...P(x_{n}=0) \\
&= \left( \frac{a+b}{2} \right) \left( \frac{2-a-b}{2} \right) \left( \frac{a+b}{2} \right) \cdots \left( \frac{2-a-b}{2} \right) \\
&= \frac 1 {2^n} (a+b)^{n/2} (2-a-b)^{n/2}.
\end{align}
$$

Side note: You initially said your problem was "Find $P(X | y=1)$" but then in the next line you started trying to compute $P(y=1|X)$. I'm kind of assuming the problem really said to compute $P(y=1|X)$, which means your Bayes' Theorem approach is the right idea. But just in case I'll say: if you really did want to compute $P(X | y=1)$ then that wouldn't be a Bayes' Theorem situation at all; you could just write $P(X | y=1) = P(x_1=1|y=1) P(x_2=0|y=1) P(x_3=1|y=1) \cdots P(x_n=0|y=1)$ and go from there.
