The representations are isomorphic, if they are dual and irreducible. Given unitary representation on a Hilbert space, you are asking how many scalar products are there after identifying them.

Up to scaling, there is only at most one inner product, which makes an irreducible representation unitary.

Now why to prefer an integral over $K$?

The parabolic induced are not always irreducible, but indecomposable, e.g. $Ind_P^G |\cdotp |^{\pm 1/2}$ for $GL(2)$. To get useful information, one often restricts to $K$.
Note that duality tells you that one of them contains an irreducible invariant subspace ($-1/2$) and the other contains an irreducible quotient ($1/2$).

The Iwasawa decomposition
$$PK =G$$
suggests that you can realize $Res_K Ind_P^G V$ as a subspace of $L^2(K) \otimes V$. So the integral over $K$ comes from the Iwasawa decomposition.

It is very common to study discrete series or super cuspidal for reductive group over non archimedean field, via their restriction to closed subgroups, which are compact modulo the centrum and maximal with this property. In Lie theory, this is the theory of weights and in the nonarchimedean case one talks about types and fundamental strata. In both cases, a finite number of $K$ reps inside the restriction determines the representation uniquely.

There is a slightly alternative way of constructing discrete series via index theorem:
Atiyah, Michael; Schmid, Wilfried A geometric construction of the discrete series for semisimple Lie groups. Invent. Math. 42 (1977), 1–62. (Reviewer: P. C. Trombi), 22E45

But I can not tell you, to what extent the constructions differ.