Palindrome P₁ =   < p9 > * < p10 > * < p56 > 
Here the year 1956 is shortened to 56 and the 19 part is implicit.

Palindrome P₂ =   < p9 > * < p10 > * < p19 > * < p56 > 

Palindrome P₃ =   < p9 > * < p10 > * < p1956 > 

Till now I don't know if this puzzle is easy or hard.
If you like the challenge be my guest !

What can I bring forward to shorten the search? E.g.
Well the palindrome must be of odd length (uneven) otherwise
the factor 11 would appear but < p2 > is not part of the answer.
The palindrome itself cannot be even as 2 or a < p1 > is not allowed.
It can not start or end with digit 5 for the same reason.
But then again three primes multiplied together can never
result in an even number or ending with 5 is it not...

And what about our current year 2022 split up like this:

Palindrome P₄ =   < p20 > * < p22 > 

Maybe here we have a feasible card.