If:
$$ \sigma_t $$ is a bounded function of both time and sample path, show that: $$ dX_t=\sigma_tX_tdW_t $$
is a P Martingale.
*Does this question make sense, that is, should the question be: is the process that solves this SDE a P martingale instead. Further, how do the characteristics of sigma help answer the question?
Please be warned that I haven't done these types of calculations in a while, so I might be missing something obvious. Anyway, this is how I would proceed.
The question makes sense as is. If we write $$ \left|\sigma_t(X_t)\right|\le M, $$ then the volatility term $|\sigma_t(X_t)|X_t$ is bounded by $MX_t$ which is sublinear and Lipschitz. Thus the solution to the stochastic differential equation exists and is unique, as long as you specify its initial value.
Denoting by $(X_t)$ the unique solution, since it solves $$ X_t=X_0+\int_0^t\sigma_t(X_s)X_s\,\mathrm dW_s, $$ it follows that $(X_t)$ is a local martingale. A sufficient condition for it to be a martingale is that for all $t\ge0$, $$ \mathbb E\left[\langle X\rangle_t^2\right]<+\infty. $$ In other words, we need to verify that $$ \mathbb E\left[\int_0^tX_s^2\sigma_s(X_s)^2\,\mathrm ds\right]<+\infty. $$ Next, note that $$ \mathbb E\left[X_t^2\right]=X_0^2+\mathbb E\left[\int_0^tX_s^2\sigma_s(X_s)^2\,\mathrm ds\right] \le X_0^2+M^2\int_0^t\mathbb E\left[X_s^2\right]\,\mathrm ds. $$ Hence by Grönwall's inequality, $$ \mathbb E\left[X_t^2\right]\le X_0^2+X_0^2M^2\int_0^te^{M^2(t-s)}\,\mathrm ds=X_0^2e^{M^2t}, $$ and therefore, $$ \mathbb E\left[\int_0^tX_s^2\sigma_s(X_s)^2\,\mathrm ds\right]\le X_0^2M^2\mathbb \int_0^te^{M^2s}\,\mathrm ds=X_0^2\left(e^{M^2t}-1\right)<+\infty. $$