= Solution
First take $0<p<1$ and $kp>1$. Part (a) gives an <extinction> <probability> $\eta\in(0,1)$. Write $p_j=\mathbb P(Z=j)$. Conditional on the ancestor having $j$ children, <extinction> requires <extinction> of all $j$ <independent> descendant trees and has <probability> $\eta^j$. Bayes' rule therefore gives the conditional offspring law
$$
\widetilde p_j=\mathbb P(Z=j\mid\text{extinction})=p_j\eta^{j-1}.
$$
Its sum is $f(\eta)/\eta=1$. Conditional on <extinction> of all descendant trees, those trees remain <independent>, each with its own extinction-conditioned law. Thus the entire conditioned process is a Galton–Watson process with offspring generating function
$$
\widetilde f(s)=\frac{f(\eta s)}\eta.
$$
For binomial offspring, using $\eta=(1-p+p\eta)^k$ gives
$$
\boxed{\widetilde Z\sim\operatorname{Bi}(k,\widetilde p),\qquad
\widetilde p=\frac{p\eta}{1-p+p\eta}.}
$$
Its <mean> is $k\widetilde p=f'(\eta)<1$. Indeed, $f(s)-s$ is <strictly convex> and has its two zeros at $\eta$ and one, so its <derivative> at the first zero is strictly negative. This proves subcriticality directly from the generating function.
Let $T$ be <total progeny>, taking value infinity on survival. Apply part (b) to the conditioned process. For some $c>0$,
$$
\boxed{1-\eta\leq\mathbb P(T>n)
=(1-\eta)+\eta\,\mathbb P(\widetilde T>n)
\leq1-\eta+\eta e^{-cn}.}
$$
Thus the unconditional tail tends to the positive survival <probability> $1-\eta$, with an exponentially small finite-extinction contribution. It does not decay to zero.
If $p=1$ and $kp>1$, <extinction> has <probability> zero and conditioning on it does not specify a <probability> law. In that deterministic case <total progeny> is infinite and $\mathbb P(T>n)=1$ for every finite $n$.
The nondegenerate law above is the <binomial branching process conditioned on extinction>.
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