We can written $a_{n}, g_{n}$ as $a_{n} = a_{1} + (n - 1)d$ and $g_{n} = k^{n - 1}.g_{1}$ We have $a_{2} = g_{2} \iff a_{1} + d = kg_{1} \implies d = (k - 1)g_{1}$ and $a_{10} = g_{3} \iff a_{1} + 9d = k^{2}.g_{1} \implies 9d = (k^{2} - 1)g_{1}$ since $g_{1} = a_{1}$ Case 1. $k = 1$ we implies that $d = 0$. It means $a_{i} = a_{1} = g_{1} = g_{p}$ Case 2. $k \neq 1$, we get $k = 8$ and $d = 7a_{1}$ Now, $g_{p} = 8^{p - 1}a_{1}$. We choose $m$ such that $a_{m} = 8^{p - 1}.a_{1} \iff a_{1} + (m - 1)7a_{1} = 8^{p - 1}a_{1} \iff 7m - 6 = 8^{p - 1} \iff m = \frac{8^{p - 1} + 6}{7}$. Note that $\frac{8^{p - 1} + 6}{7} \in \mathbb{Z}$ since $8^{p - 1} + 6 \equiv 1 + (-1) \equiv 0\pmod{7}$. Thus, we can always choose $m$ satisfy the condition.

Write $a_1 = g_1 = k$ and let the common difference and ratio of $(a_n)$ and $(g_n)$ be $d$ and $r$ respectively. We note that $$r = \frac{a_{10}}{a_2} = \frac{a_2}{a_1} = \frac{k+9d}{k+d} = \frac{k+d}{k}$$and from this we get $k(k+9d) = (k+d)^2$ which simplifies to $7k = d$, as $d$ is nonzero. From this $r = 8$. The sequence $(g_n)$ has terms $k, 8k, 64k, \ldots, 8^{p - 1}k, \ldots$ while $(a_n)$ has terms $k, 8k, 15k, \ldots, 7(m - 1)k + k, \ldots$. It remains to show that $8^{p-1} \equiv 1 \pmod{7}$, which is vacuous.