You're multiplying 1000 times by something that's equal to either $u$ or $d$ each time. That means you're adding something equal to $\log u$ or $\log d$ 1000 times. The central limit theorem applies to independent random variables that are added; that's why we're taking logarithms.
We have $ \mathbb{E}\left.\begin{cases} \log 1.1 & \text{with probability } 0.49 \\ \log 0.95 & \text{with probability } 0.51 \end{cases}\right\} = 0.49\log 1.1 + 0.51\log 0.95, $ and $ \operatorname{var}\left.\begin{cases} \log 1.1 & \text{with probability } 0.49 \\ \log 0.95 & \text{with probability } 0.51 \end{cases}\right\} = (0.49)(0.51)(\log 1.1 - \log 0.95)^2. $ Hence $ \frac{\left.\begin{cases} \log 1.1 & \text{with probability } 0.49 \\ \log 0.95 & \text{with probability } 0.51 \end{cases}\right\} - (0.49 \log1.1+0.51\log 0.95)}{\sqrt{(0.49)(.051)(\log1.1-\log0.95)}} $ has expected value $0$ and standard deviation $1$.
Adding up $1000$ independent copies of this, we get $ \frac{\text{sum}-1000(0.49\log1.1+0.51\log0.95)}{\sqrt{1000(0.49(0.51)(\log1.1-\log0.95)}} $ has an approximately standard normal distribution. The problem is: what is the probability that this is more than $ \frac{\log 1.5 - 1000(0.49\log1.1+0.51\log0.95)}{\sqrt{1000(0.49(0.51)(\log1.1-\log0.95)}}. $ (I.e. going up at least $50\%$ is the same as multiplying by $1.5$ or more.)
So that last item is what you put into the table of values of the c.d.f. of the standard normal distribution.