I take the main diagonal to mean the diagonal from $(0,n)$ to $(n,0)$. We know that it takes exactly $n$ steps until the particle is at a position $(i,n-i)$ $0\leq i\leq n$. This is because after $k\leq n^2$ steps the particle must be at a position $(j,k-j)$ for $0\leq j\leq k$ since each step increases the first coordinate position by one or the second coordinate position by one but not both.

Firstly, I take the main diagonal to mean the diagonal from $(0,n)$ to $(n,0)$. We know that it takes exactly $n$ steps until the particle is at a position $(i,n-i)$ $0\leq i\leq n$. This is because after $k\leq n^2$ steps the particle must be at a position $(j,k-j)$ for $0\leq j\leq k$ since each step increases the first coordinate position by one or the second coordinate position by one but not both.

Secondly take the main diagonal to be from $(0,0)$ to $(n,n)$. Let $N_k\sim\text{Bin}(k,p)$ be the number of up steps after $k$ steps. For the particle to be at the a point $(i,i)$ we must have $N_k=i$ and a total of $2i$ steps must have been taken. Thus the probability that after $k$ steps the particle is on the diagonal is $0$ if $k$ is odd and \begin{equation} \binom{k}{k/2}p^{\frac{k}{2}}(1-p)^{\frac{k}{2}} \end{equation} if $k$ is even.