The eleven postulates are sufficient to prove 3.11.
Lemma 1 A line and a point not on it, two different lines in a plane, or two parallel lines define a plane.
Two points on a line and a point not on it define a plane by #7. If two lines are different there's a point on the second that's not on the first (by #6), so by the first part they define a plane. By definition two parallel lines are different lines in a plane so define it by the second part.
Lemma 2 If $a,b,t$ are different coplanar lines and $a$ is parallel to $b$ and $t$ is not parallel to $a$ then $t$ is a transversal of $a$ and $b$.
By definition $t$ intersects $a$ so call the point of intersection $A$ defining an angle $\angle at\ne 0$ (by #3). Let $S$ be a point on $b$ then $SA$ defines a line $s$ (by #6) which is a transversal of $a$ and $b$ (by definition). Then $s$ cuts off angles $\angle sb=\angle sa$ (by #10) and $\angle st\ne \angle sa$ (by #4 because they are coincident), so $t$ is not parallel to $b$ by $\angle st\ne \angle sb$ and #10, and is a transversal (by definition).
Proposition If $a,b,c$ are different lines with $a$ parallel to $b$ and $b$ parallel to $c$ then $a$ is parallel to $c$.
If the lines are coplanar then let $t$ be a line intersecting $b$, then applying Lemma 2 twice it is a common traversal of $a,b,c$. By #10 $\angle ta=\angle tb=\angle tc$ and by #11 $a$ is parallel to $c$.
If the lines are not coplanar, then let $C$ be a point on $c$. By Lemma 1 $a$ and $b$ are in a plane $\pi_1$, $b$ and $c$ are in a different plane $\pi_2$, and $a$ and $C$ are in a plane $\pi_3$. By #9 $\pi_2$ and $\pi_3$ intersect in a line $l$ that contains $C$.
$l$ cannot intersect $b$ in any point $B$, otherwise $a$ and $B$ are in both $\pi_1$ and $\pi_3$, so $\pi_1\equiv\pi_3$ by Lemma 1, $b\equiv \pi_1\cap\pi_2\equiv\pi_3\cap\pi_2\equiv l$ which would require $C$ to be on $b$, contradicting that $b$ and $c$ are parallel. So $l$ does not intersect $b$ but it intersects $c$ at $C$. Since $b,c,l$ are coplanar in $\pi_2$, by Lemma 2 they cannot all be different, so $l\equiv c$.
$l$ cannot intersect $a$ in any point $A$, otherwise $b$ and $A$ are in both $\pi_1$ and $\pi_2$, so $\pi_1\equiv\pi_2$ by Lemma 1, contradicting that $a,b,c$ are not coplanar. Since $l\equiv c$ and $a$ are both in $\pi_3$ and do not intersect it follows that $a$ is parallel to $c$.
Best Answer
Using Isosceles Triangle Theorem $\overline{AD} \cong \overline{PD} \Rightarrow \angle A \cong \angle APD$. Consider $\overleftrightarrow{DP}$ and $\overleftrightarrow{CB}$. Consider the line $\overleftrightarrow{AB}$ which is a transversal to $\overleftrightarrow{DP}$ and $\overleftrightarrow{CB}$.
Using the converse of Alternate Interior Angle Theorem we see that $\angle APD \cong \angle ABC \Rightarrow \overleftrightarrow{DP}\parallel \overleftrightarrow{CB}$.
Isosceles Triangle Theorem
Let $\triangle ABC$ be a triangle such that $\overline{AC} \cong \overline{BC}$. Then $\angle A \cong \angle B$.
Proof:
Consider the correspondence $A-B-C\leftrightarrow B-A-C$, from which we can deduce that $\overline{AC} \cong \overline{BC}$, $\angle C \cong \angle C$ and $\overline{BC} \cong \overline{AC}$. By SAS Theorem $\triangle ABC \cong \triangle BAC$, therefore, because of "parts of congruent triangles are congruent" $\angle A \cong \angle B$.
Converse of Alternate Interior Angle Theorem
Consider the figure shown in the problem as a general case. If $\angle APD \cong \angle ABC$, then $\overleftrightarrow{DP} \parallel\overleftrightarrow{CB}$.
Proof:
Suppose that $\overleftrightarrow{DP} \nparallel\overleftrightarrow{CB}$. Therefore they will intersect at a point $M$. Now consider the triangle $\triangle PMB$. By the external angle theorem $\angle APD >\angle ABC \rightarrow\leftarrow.$ Therefore $\overleftrightarrow{DP} \parallel\overleftrightarrow{CB}$.
BTW, the question must be "...such that AD = PD..." Here is a graphical counterexample of the condition you posted.