Why is the Planck function continuous and not discrete? If we imagine a object made up of Hydrogen gas that is optically thick to all radiation, and is in thermal equilibrium, then, microscopically, photons will be emitted and absorbed as emission/absorption lines.
However, the overall object should emit radiation according to Planck’s Law, which describes intensity as a continuous function of wavelength (and temperature).
How does this occur and where do the photons we detect at wavelengths between spectral lines of hydrogen come from originally?
 A: Hydrogen gas is not a Hydrogen atom. During some atomic collisions one has more complicated systems than just a Hydrogen atom, i.e. they have more complicated emission/absorption spectrums. As well, there is a Doppler broadening of lines that smears sharp lines. But the principal reason is, of course, (temporarily) creating complex many-atomic (and even plasma) configurations with richer spectrum. In a thick object there is sufficient number of such complex systems.
A: I think that there is a simple contradiction in the question: either you define your object as a black body that absorbs and emits every wavelength or you talk about spectral lines.
Hydrogen is not a black body. If you imagine it so then don't be surprised if you reach contradictory conclusions.
A: You probably mean a gas of hypothetical atoms which can only interact with photons at one given frequency, i.e. atoms with the only one excitation energy level. 
If such excited atom has finite lifetime, which looks required, this automatically means it's energy level have non-zero width. And this means in turn, that absolutely cold gas will emit one non-zero width line. 
While going into equilibrium with radiation, such atoms become to move as result of photon propulsion. The more propulsion acts, the more chaotic movements. Finally such gas will be in equilibrium with radiation only when it's Doppler widening of emission lines reach the situation when all widen lines overlapping will coincide with Planck spectrum.  
A: In order to be a perfect blackbody an object must be opaque (optically thick) to radiation at all wavelengths.
If you hypothesise a material that only absorbs/emits at discrete wavelengths then this can never emit a blackbody spectrum since there is no mechanism to emit radiation outside of those discrete lines.
In practice of course, atomic hydrogen gas at any non-zero temperature will have some opacity at all wavelengths - for example there is a non-zero probability of any atom becoming ionised and then there is a possibility of recombination radiation  when a free electron combines with a hydrogen atom to form an H$^{-}$ ion$^{\dagger}$ or bremsstrahlung if it accelerates in the electric field of a proton. The inverse processes provides continuum opacity. Or, a tiny, but non-zero opacity is provided by the wings of the atomic transition lines - either because of natural line broadening or Doppler and pressure broadening. Whether these could result in blackbody radiation would depend on whether the hydrogen was physically thick and dense enough to be optically thick at all relevant wavelengths.
$\dagger$ This mechanism and its reverse is largely responsible for the continuum opacity in the solar photosphere and why the Sun's spectrum can be approximated (rather poorly actually) as a blackbody spectrum.
