= Solution
Direct multiplication gives $T^2=P^2=I$, $TP=PT=-I$, and
$$
TB(x)T=PB(x)P=B(-x).
$$
Thus conjugation by each generator of $G'$ preserves $G$; so does conjugation by every product of the generators. Therefore $\boxed{G\trianglelefteq G'}$.
Using $TB(x)=B(-x)T$ and its analogue for $P$, every group word can be reduced to $B(x)S$ with $S\in\{I,T,P,TP\}$. Consequently the quotient has at most four <cosets>. They are distinct: a nonidentity diagonal matrix among $T,P,-I$ cannot equal $B(x)$, since a diagonal $B(x)$ requires $\sinh x=0$ and hence $B(x)=I$. Therefore
$$
G'/G=\{G,TG,PG,TPG\}.
$$
The three nonidentity <cosets> all square to $G$, and $TG$ and $PG$ commute because $T,P$ commute. Hence
$$
\boxed{G'/G\cong C_2\times C_2,\quad\text{the Klein four-group}.}
$$
Equivalently, $G'$ is the full real $O(1,1)$ group and $G$ its <identity component>. This is the <component quotient of the real Lorentz group in two dimensions>; the quotient records the four components, not the continuous boost parameter.
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