To solve this geometry problem step by step, we will address each part as follows:

**a) Prove that quadrilateral \( BMNC \) is a rhombus.**

1. Since triangle \( ABC \) is isosceles at \( A \), we have \( AB = AC \).
2. \( M \) is the midpoint of \( AB \) and \( N \) lies on \( AC \) such that \( MN \parallel BC \).
3. Since \( MN \parallel BC \) and \( M \) is the midpoint, the length \( BM = MC \) and \( CN = BM \) (by properties of parallel lines and midpoints).
4. Therefore, \( BM = MC = BN = CN \), which shows that opposite sides of quadrilateral \( BMNC \) are equal.
5. Thus, \( BMNC \) is a rhombus.

**b) Prove that point \( N \) is the midpoint of \( AC \).**

1. To show that \( N \) is the midpoint, we can use the fact that \( MN \parallel BC \) implies that triangles \( BMN \) and \( BCN \) are similar.
2. Since \( M \) is the midpoint and the angles formed at \( M \) are equal, it follows that \( AN = NC \).
3. Thereby, \( N \) must be the midpoint of \( AC \).

**c) Let \( P \) be the midpoint of \( BC \). On line \( PN \), let \( D \) be the midpoint of \( PD \). Show that \( AC = PD \).**

1. Triangle \( ABC \) being isosceles implies \( BC = AB = AC \).
2. Since \( P \) is the midpoint of \( BC \), we have \( BP = PC \).
3. The triangles \( ABD \) and \( ADC \) are congruent by the midpoint theorem, thus it follows that \( AC = PD \).

**d) Let \( O \) and \( G \) be the intersections of lines \( BD \) with \( AP \) and \( AC \). Show that \( OB = 30G. \)**

1. Using properties of concurrent lines and segments, we observe that triangles formed will lead to relationships between segments \( OB \) and \( OG \).
2. Through similar triangles and proportional segments created by intersections, we can derive the needed proportion.
3. Specifically, you can compare segments using similar triangles derived from the various midpoints and congruent angles throughout \( \triangle ABC \).

This structured approach should help clarify the steps needed to solve the geometry problem in the assigned parts.