= Solution
The auxiliary <Higman group> $Q$ has no nontrivial <finite quotients of a group>. One can use this permitted property directly, but there is a short verification. In any finite quotient, let $n_i$ be the order of the image of $a_i$. Since conjugation sends $a_{i+1}$ to its square, $n_{i+1}$ must be odd. If some $n_i>1$, choose the least <prime number> $p$ dividing any $n_i$. It is odd. Conjugation by the preceding generator gives
$$
2^{n_{i-1}}\equiv1\pmod p.
$$
The <multiplicative order> of $2$ modulo $p$ is therefore a divisor of $n_{i-1}$. It is greater than one and divides $p-1$ by the <Fermat little theorem>, so it has a prime divisor smaller than $p$. That prime also divides $n_{i-1}$, contradicting the choice of $p$. Hence every $n_i=1$ and the quotient is trivial.
Given any finitely presented group $P$, form $\widehat P=P*\langle v\rangle$, and apply part (a) to the word $w=v$. This word is nontrivial by the <normal form theorem for a free product>, so the output $J=\widehat P(v)$ contains $P$ by the embedding already proved.
To check its finite quotients, let $f:J\to F$ be any <group homomorphism> to a <finite group>. The image of the copy of $Q$ is trivial by the preceding property. The identifications $u_j=a_3^{j+1}a_1a_3^{-(j+1)}$ make all $f(u_j)$ trivial. Thus the equations $u_j^{-1}cu_j=g_j$ give, in the image,
$$
f(c)=f(z)=f(d)=f(x_i z)\quad\text{for every input generator }x_i.
$$
Since $f(c)=f(z)$, this forces $f(x_i)=1$ for all generators of $\widehat P$, including $v$. The defining word $c=[v,z]$ then has trivial image, so $f(z)=f(d)=1$ as well. Every generator of $J$ has trivial image. Therefore
$$
\boxed{P\hookrightarrow J,\qquad J\text{ is finitely presented and has no nontrivial finite quotients}.}
$$
This uses the concrete construction of part (a), not an assumption that arbitrary quotients preserve an embedding of $P$.
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