Those are different situations.
In the first case, the ladder is leaning against a vertical wall. (it's not shown in your picture, but implied). The wall would then provide a normal force only in the horizontal direction.
In the second case, the ladder is leaning over a corner. In that case, the contact has a normal force based on the angle that the ladder comes into contact with the corner.
Without friction, the reaction force is always perpendicular to the contact surface, which is different in each situation.
In general, there is no reason why all the reaction forces can be determined in a problem. Your intuition is biased by all the exercises you were assigned in an academic setting, which were hand-picked to be fully solvable. In this example, intuitively, you have three equations for four unknowns so modulo some mathematical caveats, you cannot hope to find a unique solution.
Essentially, what this means is that your model is incomplete, you'll need additional information. Typically, in a realistic setting, you would say that $f_1,f_2$ arise from Coulomb friction, so the problem would still be under determined and multiple equilibrium positions exist (you can easily check it in real life). The simplest, fully solvable setting would be to assume that $f_1,f_2$ result from two springs. For notations, let's take the coordinate system to be directed by the floor ($x$) and wall ($y$), the rigidity of the spring at $P$ as $k_1$ linking it to the equilibrium point at $(x_0,0)$ and for the spring at $Q$ (the opposing extremity) as $k_2$ linking it to the equilibrium point $(0,y_0)$.
The problem is now easily solvable. The fastest route is by an energy method:
$$
E = \frac{k_1}{2}(L\cos\theta-x_0)^2+\frac{k_2}{2}(L\sin\theta-y_0)^2+mg\frac{L}{2}\sin\theta
$$
from which you can calculate $\theta$ by minimizing $E$ and deduce the four forces using:
$$
f_1 = k_1(x_0-L\cos\theta)\\
f_2 = k_2(y_0-L\sin\theta) \\
N_1 = -mg-f_2 \\
N_2 = -f_1
$$
Notice that the final forces will depend crucially on $k_1,k_2,x_0,y_0$, which means that these extra parameters were needed to fully solve the problem. In particular, you should retrieve the Coulomb friction when $k_1,k_2\to \infty$ and $x_0,y_0$ the initial position of the beam, you'll notice that the indeterminacy is caused by the dependence on the ratio $k_1/k_2$ which is still not determined in this limit.
You can reconcile the previous argument with your intuition regarding a theoretical pressure sensor. It indicates that the reading of the pressure sensor will depend on the physical characteristics of the sensor, not only on the situation. In general, the fact that you can "cancel out" the influence of the measurement device is not trivial and results from a carefully chosen experimental setup.
There exists some simpler examples where you have this kind of indeterminacy. Take for example a rigid horizontal beam that rests on three rigid pillars of the same height, check out Arnold's Mathematical Methods of Classical Mechanics for more.
Hope this helps and tell me if you need more details.
Best Answer
The sketch is no entirely clear without more context. But I guess the tilted line is a rod leaning on a wall?
At the contact point between rod and floor, there is indeed a normal force upwards preventing the rod from breaking through the floor.
But there also must be a friction force sideways. If there wasn't then the rod would have slipped sideways and fallen down. Since it doesn't, then we know that a friction force is present as a reaction to just like the normal force.