Your reasoning seems correct, but you shouldn't assume the nature of processes unless stated. I find that very uncertain, and unnecessary as in your question.
From the given diagram it is clear that there is a liner relation between $T$ and $P$,
$T=kP$ $\space $(Identical to the standard equation of a straight line, $y=mx$)
The questions says that the gas is ideal, which enables you to use ideal gas equation.
Now according to you, when the temperature increases, the pressure increases but the work is zero. The flaw is this reasoning could could be understood by considering the ideal gas equation. This reasoning only holds when the change in $T$ with respect to $P$ is constant (which is the definition of the slope of an line, $k$) which leaves you with a volume, but there can be cases where the graph between $T$ and $P$ is not a straight line, and in those cases work by the gas won't be zero.
As @Ivan added in his comment, you second statement which states that the work is zero, contradicts with your first one. This is not a very clear way to look at the process indicated by the diagram.
In case of adiabatic process, it is not necessary that work will be nil if no heat exchange takes place. It just means that heat exchange is zero, it does not tell you anything about the work done by (or on) the system.
What you need here is a change of perspective, you are thinking the temperature increases when the pressure increases (which indicates compression), but instead, you can increase temperature at constant volume which will increase the system's pressure.