Simplification of a convolution-like expression
As you say, since $\varphi_x$ is an autocorrelation we have that
$$\sum_{k=-\infty}^{\infty} w(k) \varphi_\mathrm{x}(k-n-l+p) = 
(w \star \varphi)\bigr |_{n+l-p} = g(n+l-p)$$
where $g = w \star \varphi$.
So, since $\varphi_y$ is also an autocorrelation, we have
$$\begin{align}
\sum_{l=-\infty}^{\infty} g(n+l-p)\varphi_y(l-p)
&amp;= \sum_{l=-\infty}^{\infty} g(l+n-p)\varphi_y(p-l)\\
&amp;=\sum_{m=-\infty}^{\infty} g(m)\varphi_y(n-m) &amp;
\text{upon setting}~l+n-p=m,\\
&amp;= g \star \varphi_y\bigr|_n = h(n)
\end{align}$$
where $h = g\star\varphi_y = w\star\varphi_x \star\varphi_y$.
So now you are left with $\displaystyle \sum_{p=-\infty}^\infty h(n)$ which
is unbounded. Are you sure that this last sum is with respect to $p$
and not with respect to $n$?