To show that $\sum_k p_k\phi_k$ is a characteristic function where each $\phi_k$ is, it shown here that $\sum_kp_k =1$ is sufficient.
I don't understand how $X_N = \sum_k X_k1_{N=k}$ has the characteristic function $\sum_kp_k\phi_k$, where $N$ is independent of $X_k$ and are defined on the same underlying space? $$\phi_{X_N}(t)=E[\text{exp}(it\sum_k X_k1_{N=k})] = E[\Pi_k \text{exp}(it X_k1_{N=k})]=?$$
$$\begin{align} \sum_k P(N=k) \phi_k(t) &= \sum_k E(1_{N=k}) Ee^{itX_k}\\ &= \sum_k E(1_{N=k}e^{itX_k}) \tag 1\\ &= E(\sum_k 1_{N=k}e^{itX_k})\tag 2\\ &= E(\exp(it \sum_k X_k 1_{N=k})) \tag 3\\ &= \phi_{X_N}(t) \end{align}$$
$(1)$: $N$ and $X_k$ are independent
$(2)$: $\sum_k \int | 1_{N=k}e^{itX_k}| dP = \sum_k \int 1_{N=k} dP = \sum_k P(N=k) = 1 <\infty$ so switching the sum and the expectation is licit
$(3)$: $\sum_k 1_{N=k}e^{itX_k} = \exp(it \sum_k X_k 1_{N=k})$
As a side-note, this proves the following fact:
If $\phi$ is a characteristic function and $\lambda>0$, $e^{\lambda(\phi-1)}$ is a characteristic function.
Indeed consider $N$ a Poisson$(\lambda)$ r.v. and $Y$ a r.v. having characteristic function $\phi$. Let $Y_0=0$ and $(Y_n)_{n\geq 1}$ i.i.d according to $Y$. Set $X_n=\sum_{k=0}^n Y_k$. Then $\phi_{X_n}(t)=\phi(t)^n$, hence $$\sum_{k\geq 0} P(N=k) \phi_k(t) = \sum_{k\geq 0} \frac{\lambda^k}{k!} e^{-\lambda} \phi(t)^k= e^{\lambda(\phi(t)-1)}$$