Ok. You have a pretty difficult problem.
Your x-intercepts are 0 and 4. From 0 to 3, P is decreasing and reaches a minimum at 3, but P(3)=4, so the graph must be discontinuous as you have shown. But also, from 3 to 4, P is increasing from its minimum of 4, but is zero at x=4. Again, the graph must be discontinuous.
I've seen a lot of problems like this, and they're almost always continuous. If P(3) were -4 instead of 4, it would be continuous. Did you copy the problem wrong?
Anyway, I'm sure you know that:
P' positive means increasing, P' negative means decreasing
P'' positive means concave up, P'' negative means concave down
P' zero, P'' negative means a local maximum
P' zero, P'' positive means a local minimum
P' and P'' both zero means an inflection point
below 0: decreasing, concave up - yes
0: inflection point, P(0)=0 - yes
0 to 1.5: decreasing, concave down - yes
1.5: decreasing, neither concave up nor concave down - yes
1.5 to 3: decreasing, concave up - yes
3: minimum, P(3)=4 - not a minimum
3 to 4, increasing, concave up - no (need to have second discontinuity)
above 4: increasing, concave up, P(4)=0 - yes
So you should have a graph with 3 pieces. If you do actually have P(3)=-4 instead of 4, you can move the middle piece into place. Let us know if you copied the problem wrong. If not, I would ask your professor/teacher/TA - let us know what happens. Even if you don't ask, let us know what the solution set says.