How will I solve this using DP?

Since you are interested only in the number of trailing zeroes you need only to consider the powers of 2, 5 which you could keep in two separate nxn arrays. So for the array

1 2 3
4 5 6
7 8 9

you just keep the arrays

the powers of 2    the powers of 5       
0 1 0              0 0 0
2 0 1              0 1 0
0 3 0              0 0 0

The insight for the problem is the following. Notice that if you find a path which minimizes the sum of the powers of 2 and a path which minimizes the number sum of the powers of 5 then the answer is the one with lower value of those two paths. So you reduce your problem to the two times application of the following classical dp problem: find a path, starting from the top-left corner and ending at the bottom-right, such that the sum of its elements is minimum. Again, following the example, we have:

 minimal path for the 
 powers of 2                 value
 * * -                         2
 - * * 
 - - *

 minimal path for the 
 powers of 5                 value
 * - -                         0
 * - - 
 * * *

so your answer is

 * - -                      
 * - - 
 * * *

with value 0

Note 1

It might seem that taking the minimum of the both optimal paths gives only an upper bound so a question that may rise is: is this bound actually achieved? The answer is yes. For convenience, let the number of 2's along the 2's optimal path is a and the number of 5's along the 5's optimal path is b. Without loss of generality assume that the minimum of the both optimal paths is the one for the power of 2's (that is a < b). Let the number of 5's along the minimal path is c. Now the question is: are there as much as 5's as there are 2's along this path (i.e. is c >= a?). Assume that the answer is no. That means that there are less 5's than 2's along the minimal path (that is c < a). Since the optimal value of 5's paths is b we have that every 5's path has at least b 5's in it. This should also be true for the minimal path. That means that c > b. We have that c < a so a > b but the initial assumption was that a < b. Contradiction.

Note 2

You might also want consider the case in which there is an element 0 in your matrix. I'd assume that number of trailing zeroes when the product is 1. In this case, if the algorithm has produced a result with a value more than 1 you should output 1 and print a path that goes through the element 0.