= Solution
Write the <Gaussian integer> $\alpha=a+bi$, with $a,b\in\mathbb Z$, and set $R=\mathbb Z[\alpha]$. Complex conjugation preserves $R$, because $\bar\alpha=2a-\alpha$. Choose a nonzero $\beta\in P$. Then $N=\beta\bar\beta$ is a positive integer in $P$. It cannot be one because $P$ is a proper <prime ideal>. Factoring $N$ into rational primes and applying primality repeatedly gives a prime $p\in P$.
The element $\alpha$ satisfies the monic <polynomial> $F(t)=t^2-2at+a^2+b^2$. Evaluation induces a <ring homomorphism> that is a <surjection>
$$
\mathbb Z[t]/(p,F)\longrightarrow R/P.
$$
The source has $p^2$ elements by part (b), so
$$
\boxed{|\mathbb Z[\alpha]/P|\le p^2<\infty.}
$$
If $b=0$, use $F(t)=t-a$ instead and get at most $p$ elements. Moreover $R/P$ is an <integral domain>, so its finiteness makes it a field. This proves the <finite residue-field theorem for Gaussian integer orders> without incorrectly assuming that every Gaussian integer order is a <principal ideal domain>.
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