A subalgebra of $P(M, \Delta)$ is a Peano Algebra Let $M$ be any set and let $\Delta=(n_i)_{i \in I}$ be an algebraic type. Let $P=P(M, \Delta)$ be the algebra such that the generalized Peano Axioms hold:
(P1) $f_{i}(a_0,\dots,a_{n_i-1}) \notin M$;
(P2)$ f_{i}(a_0,\dots,a_{n_{i}-1}) = f_{j}(b_0,...,b_{n_{j}-1})$ only if $i=j$ and $a_{k}=b_{k}$ for all $k$;
(P3) the set $M$ generates $P$.
The question is to prove that a subalgebra of $P(M, \Delta)$ is also a Peano Algebra. Would the correct way to approach this problem is to just check if all the axioms hold for any subalgebra of $P$?
 A: Indeed, your suggestion is correct, and the whole point is to find an
appropriate $M$.
One crucial property of Peano algebras (just as in $\mathbb{N}$ with
the successor) is that each of their elements can be expressed in a
unique way. Namely,
$$
\forall x\in P\, \exists!\, n, \overline m_x, \text{ term }t_x : \overline m_x \in
M^n \land x = t_x(\overline m_x)
$$
Given $x$, the existence of such $n, \overline m_x, t_x$ is granted by
the fact that $M$ generates $P$. Uniqueness can be proved by induction
on terms using (P1) and (P2) above.
We can take the partial order induced by the “subterm” relation on
$P$: For every $x,y\in P$ let $x\prec y$ mean that $t_x(\overline m_x)$ 
occurs as a proper subterm of $ t_y(\overline m_y)$, where we construe
elements of $M$ as constants. Now, for each subalgebra
$S\subseteq P$, take
$$
M':=\{x \in S : x \text{ is }{\prec}\text{-minimal}\}.
$$ 
It is clear that $M'$ generates $S$ and that avoids
the image of every $f_i\restriction S$. Property (P2) is also easy to
verify. Hence $S=P(M',\Delta)$.
