For Gaussian random variable $\xi_t$ with mean $\mu$ and standard deviation $\sigma$, consider the random walk with initial condition $P_0=100$, such that
\begin{equation} P_t=P_{t-1}(1+\xi_t). \end{equation}
At time $t$, what is the probability that $P_t<k$? Or what is the probability that $P_t<100$? Any help would be much appreciated. Note that the expected value of $P_t$ is trivially $100(1+\mu)^t$.
First, observe that the variance of $P_t$ satisfies $$\mathsf{Var}\, P_t=\mathsf{Var}\left(100\prod_{i=1}^t (1+\xi_i)\right)=10000\left(\mathsf{E}\left[\prod_{i=1}^t (1+\xi_i)^2\right]-\mathsf{E}\left[\prod_{i=1}^t (1+\xi_i)\right]^2\right).$$ And by independence (I assume) of the increments, $$\mathsf{Var}\, P_t=10000\left((1+2\mu+\mu^2+\sigma^2)^t-(1+\mu)^{2t}\right)=m_t^2\left(\left(1+\frac{\sigma^2}{m_t}\right)^t-1\right),$$
where $m_t=100(1+\mu)^t$ is the expectation of $P_t$. Note that you can further estimate $\mathsf{Var}\, P_t$ if you have additional information on $\sigma$ and $\mu$ or if you are interested in the asymptotic behavior as $t\to\infty$.
From there you can apply Chebychev inequality, but note that it does not concentrate much. For example, in the case $\mu=0$, the variance grows like $(1+C)^t$ for some $C>0$ as $t\to\infty$. You can get corresponding anticoncentration with Paley-Zygmund inequality.