Is the following redox reaction balanced? Al^3+ + K = Al + K^+?

Reduction half-equation: Al³⁺ + 3e⁻ → Al …… [1]
Oxidation half-equation: K → K⁺ + e⁻ …… [2]

[1] + [2]×3, and cancel 3e⁻ on the both sides. The balanced equation is:
Al³⁺ + 3K → Al + 3K⁺

The reaction usually done in molten state, i.e.
Al³⁺(l) + 3K(l) → Al(l) + 3K⁺(l)

Redox "reaction"....

Al^3+(aq) + K(s) --> Al(s) + K+(aq) ...... not balanced. Chemical equations must be mass balanced and charge balanced. This reaction is easy enough to balance both the number of atoms and the charge by inspection.
Al^3+(aq) + 3K(s) --> Al(s) + 3K+(aq) .... Now it's balanced. There are the same number of atoms of each element (mass balanced) and the charges are the same (charge balance). But it is also nonsense.

If potassium metal was dropped into an aqueous solution containing aluminum ions, there would be essentially no aluminum produced and all of the potassium would violently react with water which is much more plentiful.

2K(s) + 2HOH(l) --> 2KOH(aq) + H2(g)

========== Follow up ===========

In introductory chemistry an ionic equation usually involves water as the solvent. If there is a different solvent, then the question should say that. If the reaction takes place in the molten state, then that also should be indicated, at least with state symbols. State symbols are very important and all too often they are ignored by both teachers and students. Bad idea.

I indicated the reaction as taking place in aqueous solution, and therein lies the problem. Potassium metal will react with water. Other solvents may be used. For instance, here is a reference to a similar reaction taking place in liquid ammonia (http://www.dtic.mil/dtic/tr/fulltext/u2/021492.pdf ), but the problem is that ammonia becomes a reactant and other products (not just K+ and Al) are produced.

Yes

Reduction half reaction:
Al+3 + 3 e- → Al+3

Oxidation half reaction:
K → K+1 + 1 e-
We need 3 K on both sides of the reaction.
3 K → 3 K+1 + 3 e-

Let’s add the two balanced half reactions.

Al+3 + 3 K → Al + 3 K+1