= Solution
The original PDF specifies a <geometric distribution> on $\{0,1,\ldots\}$, with probabilities $pq^k$. Its <probability generating function> is $p/(1-qz)$. By <independence>, the total original claim count $K=\sum_iN_i$ has
$$
G_K(z)=\left(\frac p{1-qz}\right)^n,
\qquad
\mathbb P(K=k)=\binom{n+k-1}{k}p^nq^k,\quad k\ge0.
$$
The coefficient follows from the <negative binomial series>. Thus \b[one representation uses independent unit-rate <exponential distribution> steps and a <negative binomial distribution> count $K$].
There is a second useful <random sum of independent claims> representation. For one policy, the <random-sum transform identity> gives
$$
M_{T_i}(u)=\frac{p}{1-q/(1-u)}=\frac{p(1-u)}{p-u}=p+q\frac p{p-u},\qquad u<p.
$$
This is the <mixture distribution> of zero with probability $p$ and a rate-$p$ <exponential distribution> with probability $q$. Consequently
$$
\boxed{T\overset d=\sum_{j=1}^B Y_j,\qquad B\sim\operatorname{Binomial}(n,q),\quad Y_j\sim\operatorname{Exp}(p),}
$$
with the <binomial distribution> count independent of the independent steps. Equality follows by multiplying the <moment-generating functions>. This is a <negative-binomial sum of exponential claims>.
When $n=2$, the three values of $B$ give
$$
\mathcal L(T)=p^2\delta_0+2pq\operatorname{Exp}(p)+q^2\operatorname{Erlang}(2,p).
$$
Thus the zero mass is $\boxed{b=p^2}$, and the conditional positive <probability density function> is
$$
\boxed{\widetilde f_T(x)=\frac{p^2e^{-px}}{1-p^2}(2q+q^2x),\qquad x>0.}
$$
Its integral is $(2pq+q^2)/(1-p^2)=1$. Equivalently the positive-component <cumulative distribution function> is
$$
\widetilde F_T(x)=1-e^{-px}\left(1+\frac{pq^2x}{1-p^2}\right),\qquad x\ge0,
$$
so $F_T(x)=p^2+(1-p^2)\widetilde F_T(x)$. For $x<0$, $F_T(x)=0$.
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