Since the prime number $p$ divides the numbers $\overline{abc} = 100a + 10b + c$ and $\overline{cba} = 100c + 10b + a$, it must also divide their difference
$$\overline{abc} - \overline{cba} = 100(a-c) + (c-a) = 99(a-c).$$
If $p$ divides $a-c$ we are done. If not, then $p$ divides $99$, so $p=3$ or $p=11$.

If $p=3$ then $\overline{abc}$ is divisible by $3$. A number is divisible by $3$ if and only if the sum of its digits is divisible by $3$, so $3$ divides $a+b+c$. Hence, $p$ divides $a+b+c$.

If $p=11$ then $11$ divides
$$\overline{abc} = 100a + 10b + c = 99a + 11b + a - b + c = 11(9a + b) + (a - b + c),$$
which implies that $a-b+c$ is divisible by $11$. In this case $p$ divides $a-b+c$.