How to prove these two distributions are statistically/computationally indistinguishable
I think your problem is in trying to insist that $k$ is fixed.  That's getting you tied up in knots.

The statement "for fixed $k$, define $D_A$ to be the distribution..." makes no sense to me, because as you have previously defined $k$, $k$ isn't fixed -- $k$ is defined to be a random variable (that is uniformly distributed on $\{0,1\}^n$).  So, it looks to me like you are contradicting yourself.

Instead, let me suggest a better way to think about indistinguishability. Think about indistinguishability as a random experiment.  There is a formal definition of indistinguishability.  In that definition, we define an experiment.  The experiment works by running some algorithms that we have fully specified, and then observing the result.  As part of this you will need to fully specify the probability space and the event you care about.  For instance, if $F,G$ are two algorithms that depend only on a key $k$, then we might define them to be computationally indistinguishable if there does not exist a polynomial-time algorithm $A$ such that

$$|\Pr[k \leftarrow \{0,1\}^n ; A(F(k)) = 1] - \Pr[k \leftarrow \{0,1\}^n ; A(G(k)) = 1]|$$

is non-negligible.  That's a standard definition of computational indistinguishability.  Notice that the event $k \leftarrow \{0,1\}^n ; A(F(k)) = 1$ (and corresponding probability space) is fully defined, by defining some algorithms and then asking about the probability that it outputs 1.  Everything can be expressed as a statement about some random variable, where the random variable is defined by executing a well-specified algorithm (i.e., in pseudocode).  There is no step halfway along that says "fix $k$" (it's not even clear to me what that would mean).

The idea here is that we want to make a statement about the probability of some event.  That event is defined in terms of running a randomized algorithm and checking its output.  This is a well-defined and precise way to define an event in some probability space.  Once you start trying to add "fix $k$" in the middle of that, we lose that property and it's not clear what you mean. 

So, stick with the standard definitions, and try to apply them as is.  Don't try to create your own definition.  Or, if you insist on it, first translate your idea to a statement about a probability of some event where the event is defined in terms of a well-specified randomized algorithm.  You should be able to write pseudocode for each step of your algorithm, without needing English (like "fix $k$").  If you try to do that for your idea, probably you'll find that at one step along the way your pseudocode assigns to $k$ (say, by picking a random value from $\{0,1\}^n$), and then $k$ is whatever it is (if it is picked randomly, then it's not fixed; it's a random variable).

For example, here is one fact that is true.  If we define $X$ by $k \leftarrow \{0,1\}^n; X \leftarrow k$ and $Y$ by $k \leftarrow \{0,1\}^n; Y \leftarrow \{0,1\}^n$, then $X,Y$ are computationally indistinguishable and statistically indistinguishable (just apply the definition and see what you get).  Same if you define algorithms $A,B$ as in your question and define $X$ by $k \leftarrow \{0,1\}^n; X \leftarrow A_k()$ and $Y$ by  $k \leftarrow \{0,1\}^n; X \leftarrow B_k()$.

However, if you give the adversary knowledge of $k$, they are no longer distinguishable.  Thus, for instance,

$$|\Pr[k \leftarrow \{0,1\}^n ; \mathcal{A}(A_k()) = 1] - \Pr[k \leftarrow \{0,1\}^n ; \mathcal{A}(B_k()) = 1]|$$

is negligible for all adversaries $\mathcal{A}$, but

$$|\Pr[k \leftarrow \{0,1\}^n ; \mathcal{A}(k,A_k()) = 1] - \Pr[k \leftarrow \{0,1\}^n ; \mathcal{A}(k,B_k()) = 1]|$$

is not.

As an aside, often we use $A$ to represent the adversary/attacker, so using $A$ to represent some other algorithm can potentialy get a little confusing.