In the picture you've just drawn, pressure is constant for the entire process. Why? Well, the forces on the piston are $PA$ and $-mg$, and they have to sum to zero for the piston not to accelerate off either up or down. $A$, $m$, and $g$ don't change, so $P$ doesn't change either. Then the only way for V to change is for T to increase. So you haven't drawn an isothermal process.
But let's pretend you did draw an isothermal process. Then $T$ is constant, so either $P$ decreases and $V$ increases or vice-versa. Let's consider what has to happen to increase $V$, decrease $P$, and keep $T$ constant.
First: if we're going to increase $V$, the gas is going to do work on the environment. So, we need to supply some heat $Q$ which is exactly equal to the work done. So we're going to heat this container during this process, and carefully control the heat to keep $T$ constant. Alternatively, we're going to perform this process VERY SLOWLY, and allow the gas time to gain heat from the environment. Second, we need the gas to expand. How do we do that? We have to decrease the external force on the piston. We either take some weight off the top of the piston (decrease m in your picture), or grab the piston and pull up. In general, you're correct in saying that the pressure of $P$ is not going to make the gas expand without us doing something.
What I'm trying to get across is that these processes don't happen spontaneously. The ideal gas law has a lot of constants that can vary in a lot of different ways; if you want a specific process to happen (e.g. adiabatic, isothermal, isobaric, etc) you have to do something VERY SPECIFIC to the gas. In this case, you need to simultaneously provide heat AND pull on the the piston to create an isothermal process. You asked why the container would expand if $P$ was decreasing; the answer is, $P$ decreased because YOU did something to expand the container.