Your suspicions are right. It's not (quite) the same angle.
If you're looking for a treatment that is perhaps a more honest fudge, consider this one. In the left hand diagram draw a (dotted) line that goes from P through M, the point midway between the slits, and extend it a bit to the left. Then drop perpendiculars from the slits (S1 and S2) to this dotted line, hitting it at T1 and T2 (say). The distance T1 to T2, you'll be able to show easily, is exactly $d \sin \theta$. Then you need to convince yourself, by looking at the long thin triangles S1PT1 and S2PT2, that S1P is approximately equal to T1P and that S2P is approximately equal to T2P. If this is too hand wavy, try using the cosine formula in both of the long thin triangles!
What I like about this method is that
(a) It brings out the essential similarity between the geometries of Young's fringes and the diffraction grating (even though the first has a viewing plane at finite distance, but the second has a viewing plane effectively at infinity). [There's another, more algebraic, method of deriving the Young's fringes formula, based on Pythagoras, that makes it seem as though Young's fringes requires a totally different treatment from the grating.]
(b) The nature of the approximation being made is quite clear in this method.
Note: The error made by taking the path difference as $d \sin \theta$ is only 1% even if R is as small as 3$d$, if I remember rightly!