RSA given d and d = p
Yes, it's broken.  Here is the approach I see:

$$p = \text{gcd}( n, r^e - r \bmod n)$$

with quite high probability, for random $r$.

This happens because $e \equiv 1 \bmod p-1$, and hence $r^e \equiv r \pmod p$ (for any $r$).  It is unlikely that $r^e \equiv r \pmod q$, and hence $r^e - r$ has $p$ as a factor, but (probably) doesn't have $q$.