Nice example. Packages, overriding, overloading, erasure, abstractness - a dream
come true. I think that B is illegal because there are two methods the same
name. different signatures, the same erasure declared in B's hierachy (and
the one in the supertype is accessible to B, since they are in the same
javac does not give an error on B, therefore this is a compiler bug.
int f(T t) in p1.A1 is package-private, so is not inherited by p2.A2. (JLS 8.4.8)
Therefore, p1.B doesn't inherit any methods, because its direct superclass p2.A2 doesn't have any. abstract int f(X t) in p1.B (I've renamed B's formal type parameter T to X) doesn't override anything.
p1.C is in the same package as p1.B, so p1.C inherits abstract int f(String t). Since p1.C is not abstract, a compile-time error should occur. As indeed it does.
I presume Gilad missed the accessibility rule that causes p1.A1::f not to be inherited in p2.A2.
Note that **EVEN IF** p2.A2 did inherit int f(T t), it is not the case that abstract int f(X t) in p1.B overrides it. Overriding requires (220.127.116.11 clause (3)) that the overridden member p2.A2::int f(T t) is declared with default access in the same package as p1.B. Not true. Accessibility strikes again.
(For completeness, given m1 = int f(T t), and m2 = abstract int f(X t), where T and X are type variables, m1 has the same signature as m2 according to 8.4.2. Therefore m1 is a subsignature of m2, and m1 and m2 are override-equivalent. Not that this matters here.)