Noting that the solution must be time harmonic, suppress the linear combination
\begin{align*}
\begin{Bmatrix}
e^{j\omega t}\\e^{-j\omega t}
\end{Bmatrix}
\end{align*}
i.e., solve the Helmholtz equation. Note that there is no dependence on \(\psi\), the azimuthal angle. Therefore \(m=0\). Also, the Hankel functions of the first kind are tossed since waves are outgoing. Let \(h^{(2)}_n \equiv h_n\). Then general solution reads
\begin{align}\label{sldkfjsdlkfjalsdfjmdkl}
p(r,\theta) = \sum_{n=0}^{\infty} A_n h_n{(kr)} P_n(\cos\theta) \,.\tag{i}
\end{align}
To satisfy the velocity boundary conditions, the linearized momentum equation must be invoked for a time harmonic velocity:
\begin{align*}
u^{(r)} &= -\frac{1}{j\omega \rho_0}\frac{\partial p}{\partial r}\\
&= -\frac{A_nk}{j\omega \rho_0} h_n'(kr) P_n(\cos\theta)\\
&=-\frac{1}{j\rho_0c_0} h_n'(kr)P_n(\cos\theta)
\end{align*}
Therefore, at \(r=a\),
\begin{align*}
-\frac{A_n}{j\rho_0c_0} h_n'(ka)P_n(\cos\theta) &= \begin{cases}
u_0, \quad \text{ for } \theta \in [0,\pi/2)\\
-u_0, \quad \text{ for } \theta \in [\pi/2, \pi]\\
\end{cases}
\end{align*}
Thus
\begin{align*}
A_nP_n(\cos\theta) &= -\frac{j\rho_0c_0}{h_n'(ka)} \begin{cases}
u_0, \quad \text{ for } \theta \in [0,\pi/2)\\
-u_0, \quad \text{ for } \theta \in [\pi/2, \pi]\\
\end{cases}
\end{align*}
To solve for the expansion coefficients \(A_n\), multiply both sides by \(P_m(\cos\theta)\sin\theta d\theta\) and integrate:
\begin{align*}
\int_{0}^{\pi} A_nP_n(\cos\theta)P_m(\cos\theta)\sin\theta d\theta &= -\frac{j\rho_0c_0 u_0}{h_n'(ka)} \bigg[ \int_{0}^{\pi/2} P_m(\cos\theta) \sin\theta d\theta
- \int_{\pi/2}^{\pi} P_m(\cos\theta)\sin\theta d\theta
\bigg]\\
A_n \frac{2}{2n+1}&= \frac{j\rho_0c_0 u_0}{h_n'(ka)} \bigg[ \int_{0}^{1} P_n(z) dz
- \int_{-1}^{0} P_n(z) dz\bigg]
\end{align*}
\(P_n(z) = P_n(-z)\) is for even \(n\), so the term in the square brackets is \(0\) in that case. Meanwhile, \(P_n(-z) = -P_n(z)\) for odd \(n\), so
\begin{align}\label{sldkfjsdlkafjlsdk}
A_n &= (2n+1)\frac{j\rho_0c_0 u_0}{h_n'(ka)} \int_{0}^{1} P_n(z) dz \tag{ii}
\end{align}
To evaluate the integral, use the property \((2n+1)P_n(z) = P_{n+1}'(z) - P_{n-1}'(z)\), which is integrated trivially using the fundamental theorem of calculus:
\begin{align*}
\int_{0}^{1}P_n(z)dz &= \frac{1}{2n+1} \int_{0}^{1} \big[P_{n+1}'(z) - P_{n-1}'(z)\big]dz\\
&= \frac{1}{2n+1}[P_{n+1}(1) - P_{n-1}(1) - P_{n+1}(0) + P_{n-1}(0)]\,.
\end{align*}
Since the Legendre polynomials are normalized such that \(P_n(1) = 1\), the first two terms above cancel, giving
\begin{align*}
\int_{0}^{1}P_n(z)dz &= \frac{1}{2n+1} [- P_{n+1}(0) + P_{n-1}(0)].
\end{align*}
The two Legendre polynomials above can be combined into one by noting that \((2n+1)zP_n(z) = (n+1)P_{n+1}(z) + nP_{n-1}(z)\), or for \(z=0\), \[-P_{n+1}(0) = \frac{n}{n+1}P_{n-1}(0)\]
Thus the integral becomes
\begin{align*}
\int_{0}^{1}P_n(z)dz &= \frac{1}{2n+1}\bigg[\frac{2n+1}{n+1} P_{n-1}(0)\bigg] = (n+1)^{-1}P_{n-1}(0)\,.
\end{align*}
and equation (\ref{sldkfjsdlkafjlsdk}) becomes
\begin{align*}
A_n &= \frac{2n+1}{n+1}\frac{j\rho_0c_0 u_0}{h_n'(ka)}P_{n-1}(0)
\end{align*}
The problem is solved by inserting the coefficients above into equation (\ref{sldkfjsdlkfjalsdfjmdkl}):
\begin{align}\tag{iii}\label{lksjdaflkj}
p(r,\theta) = j\rho_0c_0 u_0 \sum_{n=0}^{\infty} \frac{2n+1}{n+1}\frac{h_n{(kr)}}{h_n'(ka)}P_{n-1}(0) P_n(\cos\theta)
\end{align}
Taking the limits of equation (\ref{lksjdaflkj}) for \(n =1\) (the leading order term) gives
\[p = -\frac{3}{4} \rho_0c_0 u_0 (ka)^3 \frac{e^{j(\omega t - kr)}}{r} \cos\theta\,,\]
i.e., the dipole directivity pattern.