Intrinsic parity

You are viewing this the wrong way. If you apply parity operation to a scalar field (both real and complex field), you have: $$ Pphi (x) P^{-1} = n_Pphi (tilde{x}) tag{1} $$ where, $n_P$ is the intrinsic parity,

and the parity operator effects the coordinate transformation from $x$ to $tilde{x} = (t, -boldsymbol{x})$.

Now, if you apply the parity operation again, $$ P^2phi (x) P^{-2} = n_P Pphi (tilde{x}) P^{-1} = n_P^2 phi (x) $$ We should obtain the original field back. Thus, we see that $n_P^2 = 1$, which can only imply that $n_P = pm 1$.

We cannot determine intrinsic parity with free Lagrangian; for that, we need interactions. Suppose you have an interaction Lagrangian of the form, $$ L_{int} = - mu phi^3 - lambda phi^4 $$ Applying the parity transformation on the interaction Lagrangian, and subsequently using Eqn. (1) for each $phi$, we see that parity invariance would require that for the first term, $n_P^3 = 1$, and for the second term $n_P^4 = 1$. Both of these equations are satisfied simultaneously only if $n_P = 1$. Hence, for this case, $phi$ is a scalar field.