Calculating Conditional Probability in a small cipher problem

I have a small part of data that is n bit long (0 or 1). Probability that a bit is 0 equals p. I also have a key that is used to cipher this data that is 1 bit long. The key will be 0 or 1 for the whole data (equal probability). A XOR operation between the key and the data takes place, and we take the encrypted data.

If the encrypted data contains k zeros (and n-k ones), calculate the probability that the key used is 0.

I am trying to solve this. My thinking until now is that we have 2 events:

Event(A). Original data had k zeros

I can calculate through the binomial distribution.

$$P(A) = {{n}\choose{k}}(p^k(1-p)^{n-k}$$

Event(B). Encrypted data has k zeros Since a XOR operator is applied if the key was 0, we will have k bits that are zero, if the key bit was 1 we will have n-k bits that are zero. So $$P(B) = \frac{P(A)}{2}$$

And then i am searching for the conditional probability $$(P(key=0|P(B))$$

Is my thinking correct?

Let $$K$$ be the event that the key is $$1$$. We seek for $$P(K'|B)$$, where $$B$$ conforms to your definition, i.e. encrypted data has $$k$$ zeros. Via Bayes rule, we could write: $$P(K'|B)=\frac{P(B|K')P(K')}{P(B)}$$ Here $$P(K')=1/2$$, and $$P(B|K')=P(A)$$ (for $$A$$, I'm using your def. again) because when the key is $$0$$, having $$k$$ zeros means having $$k$$ zeros from the beginning. We can write $$P(B)$$ via total probability theorem: $$P(B)=P(B|K')P(K')+P(B|K)P(K)=\frac{P(B|K')+P(B|K)}{2}$$ Here, $$P(B|K)=P(n-k \ \text{zeros in the original data})={n \choose n-k}p^{n-k}(1-p)^k$$. Summing the two yields (noting that $${n \choose k}={n \choose n-k}$$): $$P(K'|B)=\frac{p^k(1-p)^{n-k}}{p^k(1-p)^{n-k}+p^{n-k}(1-p)^k}=\frac{1}{1+\left(\frac{p}{1-p}\right)^{n-2k}}$$
As you can see $$P(B)\neq P(A)/2$$, instead $$P(B \cap K')=P(A)/2$$, i.e. the numerator in Bayes formulation. Also, $$P(key=0|P(B))$$ doesn't have meaning, instead you need to write $$P(key=0|B)$$.