Funny problem! Let's suppose that $\angle BDA + \angle CAB =90$ did not hold. Let's set out to show that $\angle BAD + \angle BCD = 90.$ Let $O$ be the circumcenter of $\triangle ABD$, we then know that $O \notin AC$ by our assumption. Let $Q \in OC$ be a point satisfying $OQ \cdot OC = OD^2 = OB^2.$ Then, we've that $\angle BDQ + \angle DBQ = \angle ODB + \angle OBD - \angle ODQ - \angle OBQ = 180 - 2 \angle BAD - \angle OCD - \angle OCB$, where we used that $\triangle ODC \sim \triangle OQD, \triangle OBC \sim \triangle OQB.$ We also have that $\angle BDP + \angle DBP = 90 - \angle BAD - \angle ACD + 90 - \angle BAD - \angle ACB = 180 - 2\angle BAD - \angle BCD$ by the condition, and so therefore comparing this with the previous equality gives us that $\angle BDQ + \angle DBQ = \angle BDP + \angle DBP.$ Therefore, we've that $DQPB$ is cyclic. Now observe that $\angle QDP = \angle QDB - \angle PDB = (\angle ODB - \angle ODQ) - \angle PDB = 90 - \angle BAD - \angle OCD - (90 - \angle BAD - \angle ACD) = \angle ACD - \angle OCD = \angle ACO = \angle PCQ,$ hence implying that $QDCP$ is cyclic. Combining this with the fact that $DQPB$ is cyclic hence gives us that $DCBPQ$ is cyclic, and so in particular we have that $CBPD$ is cyclic. From here, the problem falls to some more simple angle-chasing. Indeed, observe that $BPD = 180 - \angle PDB - \angle PBD = 2 \angle BAD + \angle BCD$, and so hence $\angle BCD + \angle BPD \Rightarrow 2 \angle BAD + \angle BCD + \angle BCD = 180 \Rightarrow \angle BAD + \angle BCD = 90,$ as desired. $\square$