Suppose that $A \subseteq B$.
If $A$ is isomorphic to $P$ (with isomorphism $\phi$) and $B$ isomorphic to $Q$ (with isomorphism $\psi$), does this imply that $P\subseteq Q$?
I am thinking yes, but I am not too sure.
Let $x \in P$, then $x = \phi(z)$ for some $z \in A$. But $z \in B$ since $A\subseteq B$ and so $\psi(z) \in Q$.
But $\psi(z)|_A = \phi(z)=x$ and so, we must have that $\psi(z)=x$, i.e. $x \in Q$.
Is this argument valid?
On
No. Consider the following. Let $$A = \lbrace x\in \mathbb{C}\mid x^{4} = 1\rbrace,\quad B = \lbrace x\in\mathbb{C}\mid x^{8} = 1\rbrace$$ be multiplicative groups. Then clearly $A\subset B$. Also, $A\cong \mathbb{Z}/4\mathbb{Z}$, where here $\mathbb{Z}/4\mathbb{Z}:=\lbrace 0,1,2,3\rbrace$ with addition modulo $4$, and trivially $B\cong B.$ However, $\mathbb{Z}/4\mathbb{Z}\not\subset B.$
Your last step is invalid since you haven't assumed anywhere $\psi|_A = \phi$. Take for example $A = \Bbb Z, B = \Bbb R = Q, P = \{ \left( \begin{matrix} n & 0 \\ 0 & n \end{matrix} \right): n \in \Bbb Z \}$. Of course, $A \simeq P$ (as rings) but $P \not\subset Q$.
This answer may or may not be completely accurate because you haven't really defined which kind of objects/morphisms we're talking about, but this is the general idea.
Hope it helps.