Let's make some assumptions for the purposes of calculation. Suppose 6.7 billion people have been alive for 13.5 billion years each shuffling one deck a thousand times a second.
The total number of rounds of shuffles $N=1000 \times 3600 \times 24 \times 365 \times1.35 \times 10^{10}.$
So $N = 4.25736 \times 10^{20}.$
Let $n=52! \approx 8.0658 \times 10^{67}$ and let $m=6.7 \times 10^9.$
When everyone shuffles the cards once in the first millisecond the probability that they are all different is
$$ \prod_{i=1}^{m-1} \left( 1- \frac{i}{n} \right) \approx 1 - \frac{m(m-1)}{2n},$$
ignoring the cross products since $n$ is much bigger than $m$.
Each shuffle is independent, so after $N$ steps the probabilty that we have had no match is
$$\left( 1 - \frac{m(m-1)}{2n} \right)^N \approx 1 - \frac{m(m-1)N}{2n}, \qquad (1)$$
since $m(m-1)/2n$ is very small.
Hence the probability that there has been a match is 1 minus RHS of (1), that is
$$\frac{m(m-1)N}{2n}.$$
I make this about $1.18 \times 10^{-28}$ with the figures I've used.
REMARK: Note that if you drop the "at one moment" requirement. That is, if you now ask if two shuffles have ever been the same, the problem is slightly quicker. The total number of shuffles that have ever taken place is $mN$ and so the probability that they were all different (as in the birthday problem) is $\prod_{i=0}^{mN-1} \left( 1- \frac{i}{n} \right),$ where as before $n=52!,$ So the probability that we've had a repeat is $1- \prod_{i=0}^{mN-1} \left( 1- \frac{i}{n} \right),$ which, similar to before, is approximately $mN(mN-1)/2n,$ or $5 \times 10^{-8}.$
Note that as the index $i$ in the product gets larger, better approximations are obtained by passing to logarithms (as on the wikipedia page).