Explain: An NTRU key for a lattice of dimension $d$ has size only $O(d\log{d})$ The associated NTRU lattice is of dimension $d=2N$.
The public and private key sizes are both of length $O(N)=O(d/2)=O(d)$.

So where does the $d\log(d)$ appear?

Thanks in advance.
An NTRU public key is a polynomial of degree $d-1$ whose coefficients are elements in $\mathbb{Z}_q$.  So the number of bits you need to represent it is $d\log{q}$.  If q is of the same order as d, then this gives you $d\log{d}$.  The private key, on the other hand, can be represented with $O(d)$ bits since the coefficients are all in the set $\{-1,0,1\}$.