when will a player be unable to make a move...if it is said that a cell can have more than one coins

I recommend doing the previous problems on this chapter before doing this one. Especially the first chessboard game. If you don't have an understanding of how solving a nim game works, you should read about it before trying this problem. You will need an understanding of how to use the sprague-grundy theorem to solve this.

What is the meaning of "move optimally"?

java solution, passes all test cases in O(k) time and O(1) space. First compute the nimber (e.g. Grundy number) for all 15 x 15=225 spots on the chess board. The nimber of a spot (x,y) is the smallest non-negative integer not equal to the grundy numbers of any of the up to four spots reachable from (x,y) in a single jump. This can be done in O(m^2) time, where m=15=O(1) in this problem, by filling in successive diagonals of the matrix of nimbers.

Once we have the nimbers of every spot on the chessboard, simply take cumulative XOR of all k nimbers. If 0, the first player loses, and if nonzero, the first player wins.

import java.util.Scanner;

class Solution{
    static int[][] nimbers(final int side){
	int[][] nb=new int[side][side];
	for(int j=2;j<2*side-1;++j){
	    int i=j<side?0:j-side+1;
	    int k=j<side?j:side-1;
	    while(i<=k){
		boolean[] seen=new boolean[4+1];
		if(i>1){
		    seen[nb[i-2][k-1]]=true;
		    if(k!=side-1) seen[nb[i-2][k+1]]=true;
		}
		if(k>1){
		    if(i!=0) seen[nb[i-1][k-2]]=true;
		    if(i!=side-1) seen[nb[i+1][k-2]]=true;
		}
		int l=0;
		while(seen[l]) ++l;
		nb[i][k]=l;
		nb[k][i]=l;
		++i;
		--k;
	    }
	}
	return nb;
    }
    static boolean win(int nCoin, Scanner sc, int[][] nb){
	int nimSum=0;
	while(nCoin-- != 0){
	    int x=sc.nextInt(), y=sc.nextInt();
	    nimSum ^= nb[x-1][y-1];
	}
	return nimSum!=0;
    }
    public static void main(String[] args){
	Scanner sc=new Scanner(System.in);
	int nCase=sc.nextInt(), side=15;
	int[][] nb=nimbers(side);
	while(nCase-- != 0){
	    int nCoin=sc.nextInt();
	    System.out.println(win(nCoin,sc,nb)?"First":"Second");
	}
	sc.close();
    }
}

Too difficult for me