PC Algorithm Causal Inference What are some reasons why the PC algorithm might generate zero edges for a certain dataset? How and why could this happen?

As @ReelSaemon's answer points out, if the data is suggesting complete independence, you get a completely disconnected graph (no edges).
You asked in the comments to @ReelSaemon's answer, how it can be that adding nodes could turn a graph with edges into an edge-free graph. In general, if $G$ is a Bayesian network and $M$ is a marginalization of $G$, then if $M$ has edges, $G$ must have edges, too. If there is an edge in $M$ between two nodes $x_1$ and $x_2$, there must be an open path between $x_1$ and $x_2$ in $G$.
So there is definitely a contradiction in your results which is due to the probabilistic nature of the problem. Lots of things can go wrong, especially if the independency relations are not very clear and there is not a sufficient amount of data. And since PC is a nontrivial algorithm, it is difficult to investigate its numerical stability.
Having said that, I just want to point out one way in which this phenomenon of "edge removal" in the larger graph $G$ could happen. Let's say there is an edge in the marginalization $M$ between $x_1$ and $x_2$ and when switching to the original graph $G$, this edge becomes a collection of several connections, as shown in this figure:

Now, imagine that for each of those nodes $y_i$, the edges of the path from $x_1$ to $x_2$ via $y_i$ describe only weak dependencies, too weak to be recognized by PC. But in the marginalization $M$, only the combined strength of all the paths is measured. And this combination can get you a strong enough dependence in $G$ between $x_1$ and $x_2$. E.g., the variables $y_i$ could be a collection of weak predictors that, when boosted, get you a strong prediction of both $x_1$ and $x_2$.
The PC algorithm working on $G$ will also recognize the strong connection between $x_1$ and $x_2$ e.g. when testing the unconditional independence between $x_1$ and $x_2$ and it probably rejects independence there, too. But the edge between $x_1$ and $x_2$ will be removed when testing the independence $x_1\perp\kern-5pt\perp x_2\;| (y_1,y_2,y_3,y_4)$.
