Non-bonded orbitals in water
General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is $\mathrm{sp}^3$ hybridized.

No. One could say that, but $\mathrm{sp}^2$ is equally possible. In fact, it may seem more likely, we’ll get to that in a second.

Group Theory and Quantum Chemistry perspective: Water belongs to $C_{2\mathrm{v}}$ point group. There are four irreducible representations, and the atomic orbitals belong to these irreducible representations (or linear combinations of these irreducible representations). Hence, we get five occupied molecular orbitals of symmetries $\mathrm{a}_1, \mathrm{b}_1$, and $\mathrm{b}_2$.

Yes, and you did a nice job of drawing them.

There is only one non-bonding orbital ($1\mathrm{b}_1$) in which the AOs of hydrogens do not participate. Where is the second lone pair orbital of water?

And this is where it gets interesting.
I’ll once again point to Professor Klüfers’ web scriptum for the basic and inorganic chemistry course in Munich, section about localising molecular orbitals. If you don’t understand German, all you need to do is look at the pictures and understand that wenig means a little in this context.
You see, to do a discussion such as ‘this is the bonding electron pair’ we need to localise molecular orbitals. Specifically by linear combining them. It’s almost like in the linked pictures: $\Psi_2 (1\mathrm{b}_2) + \Psi_3 (3\mathrm{a}_1)- \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_1$. Invert the signs: $\Psi_2 (1\mathrm{b}_2) - \Psi_3 (3\mathrm{a}_1)  + \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_2$. These are your two bonding orbitals to hydrogen. We have some $\Psi_1$ remaining (and, theoretically, also some $\Psi_2$ and $\Psi_3$, since we didn’t use all of them). That (remaining linear combination) is our second non-bonding orbital.

Now why is $\mathrm{sp}^3$ a bad description? You see, we arrived at one lone pair that is definitely $\unicode[Times]{x3C0}$-shaped or p-shaped. The second is somewhat p-ish but also somewhat s-ish, so maybe some sp-hybrid. But there is no mistaking that $1\mathrm{b}_1$ is in no way influenced by all the others. It is antisymmetric to the $\ce{H-O-H}$-plain of symmetry while all the others are symmetric. $\mathrm{sp}^2$ describes this a lot better.