The two-digit palindromic numbers are 11, 22, ..., 99; among these, 11 is a prime number, and the others are multiples of 11. We will show that this is generally true, i.e., if $P$ is a palindromic number with an even number of digits, then it is divisible by 11. This implies that the only solution to the problem is 11.

If $P$ has $2n$ digits and the first $n$ digits are $a_{1}, a_{2}, \ldots, a_{n}$, then the second $n$ digits, according to the condition, are $a_{n}, a_{n-1}, \ldots, a_{1}$. Therefore, $P$ can be written as the sum of at most $n$ numbers of the form $\overline{a_{i} 00 \ldots 00 a_{i}} \cdot 10^{i-1}$, where the number of zeros in the middle is even. (If $a_{i}=0$, then the corresponding term does not appear in the sum.)

Such a number is an integer multiple of $100 \ldots 001$ (specifically, $a_{i} \cdot 10^{i-1}$ times), which can be written as $11 \ldots 11 - 10 \cdot 1 \ldots 1$. In this latter expression, the second factor being subtracted consists of 2 fewer 1s than the minuend, and thus also an even number of 1s. Since numbers consisting of an even number of 1s—being the sum of 11 times the appropriate even powers of 10—are divisible by 11, both terms in the above difference are divisible by 11, and thus the difference itself is divisible by 11.

Therefore, $P$ is a sum of multiples of numbers divisible by 11, and thus is indeed divisible by 11. This completes the proof.

Remark. It is well-known that a number is divisible by 11 if and only if the difference between the sum of the digits in the even positions and the sum of the digits in the odd positions in its decimal representation is divisible by 11—generally, the number and this difference give the same remainder when divided by 11. Using this, we immediately see that a palindromic number with an even number of digits is divisible by 11, since in this case the aforementioned difference is zero.
Let's determine the prime number $p$, knowing that it has an even number of digits, and the number is a palindrome, meaning if we write its digits in reverse order, we also get $p$.