Fortunately you cannot prove it. It's invalid.
Counterexample. Consider a model M with a single world w ∈ |M| such that there is no reflexive arrow on w (i.e. ¬wRw). That gives us the fact that: (M, w) |= ▢ P, because it is vacuously true that for all worlds v ∈ |M| accessible from w (wRv), we have it that (M, v) |= P. That allows us to conclude that (M, w) |= ▢ P. But it is not the case that there is a world v ∈ |M| which is accessible from w and is such that (M, v) |= P (you know, given that there is only a single world and it is inaccessible from itself). So we cannot conclude that (M, w) |= ♢P.
The reasoning is similar to the usual one associated with the quantifiers ∀, ∃. If you have a universally quantified formula φ (i.e. ∀φ), if you interpret it in an empty domain, it will be satisfied. But the existential will not. Note also, that even if you have non-empty domain of discourse, it might still be the case that the interpretation I of φ specifically is empty, so the universal closures of φ might still evaluate to true vacuously, while the existential ones will not (given that no object falls in the extension of φ under the I).