If $p_X(k)$ and $p_{Y|X}(y|k)$ are given, how can we calculate $p_Y(y)$? We cannot assume that $p_Y(y)$ and $p_X(k)$ are independent.
I know that $p_X(k) \cdot p_{Y|X}(y|k) = p_{Y,X}(y,k)$ but how can I isolate $p_Y(y)$?
If $p_X(k)$ and $p_{Y|X}(y|k)$ are given, how can we calculate $p_Y(y)$? We cannot assume that $p_Y(y)$ and $p_X(k)$ are independent.
I know that $p_X(k) \cdot p_{Y|X}(y|k) = p_{Y,X}(y,k)$ but how can I isolate $p_Y(y)$?
Since you are dealing with discrete random variables, the way you find $p_Y(y)$ is using summations:
$p_Y(y) = \sum_k p_{Y,X}(y,k) = \sum_k p_{Y\mid X}(y\mid k)p_X(k)$
where the sum on the right should be recognized as coming from the law of total probability. Note that $p_{Y\mid X}(y\mid k)$ is the conditional probability of the event $A = \{Y = y\}$ given that the event $B_k = \{X = k\}$ has occurred, while $p_X(k) = P\{X = k\}$ is just $P(B_k)$. The law of total probability tells us that
$P(A) = \sum_k P(A\mid B_k)P(B_k)$
where the events $B_k$ are a (countable) partition of the sample space.
Hint:
Use $p_Y(y)=\int_x p_{Y,X}(y,x) dx = \int_x p_X(x) \cdot p_{Y|X}(y|x) dx$
EDIT:
For discrete case, just replace integration by summation as pointed out by Dilip.
P.S: How come you knew that he meant discrete and not continuous random variable??