My synthesized compound is 3,4-dihydropyrimidone derivative, which has 2 nitrogen in the ring. The $\ce{NH}$ protons are not appearing in the H NMR spectrum and the solvent used was deuterated chloroform $(\ce{CDCl3})$. I understand that the peak for $\ce{NH}$ proton tends to be broad, is there a possibility that it merges/overlaps with the neighboring peaks?
As indicated by @Waylander, your compound may undergo keto-enol tautomerism. Beside a sample which may contain varying concentrations of three compounds (top row), the products of an intentional proton-deuterium exchange equally may be recognizable (bottom row), too:
Especially in humid climate/summer season, bottles of $\ce{CDCl3}$ catch water with each opening of the vial, the $\ce{H2O}$ peak in the NMR spectrum becomes more prominent. Note, Wikipedia's property box reports for $\pu{20 ^\circ{}C}$ a solubility of $\pu{8.7 g}$ of chloroform per litre of water. An additional potential cause is storage of NMR solvents in a fridge; if you lift the lid of the still cold bottles too early and draw the liquid with a Pasteur pipette (before the bottle warmed to ambient temperature), water merely condensates on them.
Possible solutions:
The sequence represents an increase in financial investment per sample for the deuterated solvent. On the other hand, the substitution of $\ce{CDCl3}$ by DMSO-$d_6$ may save time to record and subsequently interpret the NMR spectra.
My synthesized compound is 3,4-dihydropyrimidone, which has 2 nitrogen in the ring. The $\ce{NH}$ protons are not appearing in the H NMR spectrum and the solvent used was deuterated chloroform $(\ce{CDCl3})$. I understand that the peak for $\ce{NH}$ proton tends to be broad, is there a possibility that it merges/overlaps with the neighboring peaks?
I know Buttonwood's answer is good enough to explain why you didn't see your $\ce{NH}$ peaks. I like first two of his solutions. However, as suggested in third solution, I wasn't a fan of using $\ce{DMSO}$-$d_6$ if I have a choice between $\ce{CDCl3}$ and $\ce{DMSO}$-$d_6$ because $\ce{DMSO}$-$d_6$ is highly hydroscopic than that of $\ce{CDCl3}$ (in addition to the financial setback).
When I analyzed your $\mathrm{^1H}$-NMR you have just posted, I have seen massive water peak around $\pu{1.56 ppm}$. If you have integrated that peak, you'd see the integration value is more than 3-protons of the real compound. That means your compound is really wet when you take the NMR. The following diagram shows ratios of water peak to TMS and residual $\ce{CHCl3}$ peaks in a typical $\ce{CDCl3}$ (with 99.8% $\ce{D}$ and 0.05% TMS) solvent bottle (Ref.1):
Assuming you have used similar $\ce{CDCl3}$ solvent bottle, your water peak is way bigger than expected. That means, you have either didn't dry your product (assuming this is crude product) or your $\ce{CDCl3}$ solvent bottle is really old absorbing way over water by time (as suggested by Buttonwood). Either way, this water is capable of exchanging acidic protons (similar to $\ce{D2O}$ shake) in the product so that equilibrium is moving water direction since presence of water is too much (Le Chatelier principle). Thus, you see only water molecules instead of two $\ce{N-H}$ peaks.
Other than that, your compound is pretty pure. If it is a solid, one more recrystallization would do the trick.
Note: To find out your sample is wet or your $\ce{CDCl3}$ solvent bottle is wet, just run a NMR of the solvent. If water peak is still more than TMS peak, it is really wet.
Late Edit: For my surprise, most of NMR of the derivatives of sought compound have been done in $\ce{DMSO}$-$d_6$ solvent. Following is $\mathrm{^1H}$-NMR of your exact compound in $\ce{DMSO}$-$d_6$ (Ref.2). You may see both $\ce{N-H}$ peaks clearly. So, I think you should run another spectrum in $\ce{DMSO}$-$d_6$ to confirm it:
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