As the other answer says, it is true for a simplified model. The model you describe has only one ligand binding site per protein, which makes it the most simple model there is.
It doesn't mean that this is the case at equilibrium as you asked. This is because Kd is not necessarily equal to [L] at equilibrium. Hopefully this simple derivation will help:
As you said, we know the formula for the dissociation constant: Kd = [P][L]/[PL]
Suppose [L] = Kd. If we plug back into the equation above we get: [L] = [P][L]/[PL]
Doing some algebra we can derive: [P] = [PL]
Therefore, at equilibrium, when [L] = Kd, the
amount of free protein ([P]) and the amount of ligand-bound protein ([PL]) are equal. In other words, half of the protein is bound to ligand, which is exactly what you were asking about.
But now let's suppose that [L] =/= Kd. The reaction will still come to equilibrium, finding concentrations of [P] and [PL] that satisfy the equation Kd = [P][L]/[PL]. However, these values will not be equal. Either more or less than half of the protein will be bound to ligand.
Only in the specific case of [L] = Kd will half of the protein be bound to ligand when equilibrium is reached.