A three-phase two-winding transformer has a voltage transformation ratio VP/VS = 0.866 + j0.5, where VP is the primary side voltage in p.u., and VS is the secondary side voltage in p.u. I_P and I_S represent the currents injected into the primary and secondary sides of the transformer, respectively. The admittance corresponding to the leakage impedance of the transformer referred to the secondary is yt p.u. Neglect the magnetizing branch. The Y bus representation of this transformer is:

The correct admittance matrix (Y-bus) representation for the given three-phase two-winding transformer is:
$$\left[\begin{array}{c}{I}_P\\ I_S\end{array}\right]=\left[\begin{array}{cc}{y}_t& -\frac{y_t}{0.866-j0.5}\\ -\frac{y_t}{0.866+j0.5}& y_t\end{array}\right]\left[\begin{array}{c}{V}_P\\ V_S\end{array}\right]$$

Step-by-Step Derivation and Analysis:

1. Identify the Transformer Parameters and Diagram Labels:
From the provided question and diagram (where primary side terminal is labeled as $P$ and secondary side terminal is labeled as $S$):
- The voltage transformation ratio is defined as:
$$a=\frac{V_P}{V_S}=0.866+j0.5$$
- The magnitude of this complex turns ratio is:
$$\left|a\right|=\sqrt{{0.866}^2+{0.5}^2}=\sqrt{0.75+0.25}=1.0$$
- The complex conjugate of the turns ratio is:
$$a^{*}=0.866-j0.5$$
- The series admittance corresponding to the leakage impedance referred to the secondary side is $y_t$.
- The primary and secondary injected currents are $I_P$ and $I_S$, and the terminal voltages are $V_P$ and $V_S$, respectively.

2. Mathematical Modeling of the Transformer:
An off-nominal phase-shifting transformer with complex ratio $a:1$ and leakage admittance $y_t$ referred to the secondary side can be modeled with an ideal transformer in series with the series admittance.
Let $V_S^{\text{'}}$ be the internal voltage on the secondary side of the ideal transformer. The ideal turns ratio gives:
$$V_S^{\text{'}}=\frac{V_P}{a}$$
The admittance $y_t$ is connected between this internal node $V_S^{\text{'}}$ and the external secondary terminal $V_S$. The current flowing from the internal node to the secondary terminal is:
$$I_{\mathrm{series}}=(V_S^{\text{'}}-V_S)y_t=(\frac{V_P}{a}-V_S)y_t$$
Since $I_S$ is the current injected into the secondary node, we have:
$$I_S=-I_{\mathrm{series}}=-\frac{y_t}{a}{V}_P+y_t{V}_S$$

3. Relate to the Primary Current:
By conservation of complex power in the ideal transformer portion:
$$V_P{I}_P^{*}+V_S^{\text{'}}(-I_{\mathrm{series}}{)}^{*}=0$$
Substituting $V_S^{\text{'}}=\frac{V_P}{a}$:
$$V_P{I}_P^{*}=\frac{V_P}{a}{I}_{\mathrm{series}}^{*}$$
Taking the complex conjugate of both sides:
$$V_P^{*}{I}_P=\frac{V_P^{*}}{a^{*}}$$
Since $V_P\ne 0$, dividing by $V_P^{*}$ yields:
$$I_P=\frac{I_{\mathrm{series}}}{a^{*}}=\frac{1}{a^{*}}(\frac{V_P}{a}-V_S)y_t=\frac{y_t}{a{a}^{*}}{V}_P-\frac{y_t}{a^{*}}{V}_S$$
Since the magnitude squared $a{a}^{*}=|a{|}^2=1.0$, the primary current equation simplifies to:
$$I_P=y_t{V}_P-\frac{y_t}{a^{*}}{V}_S$$

4. Construct the Matrix:
Putting the nodal equations for $I_P$ and $I_S$ together:
$$I_P=y_t{V}_P-\frac{y_t}{0.866-j0.5}{V}_S$$
$$I_S=-\frac{y_t}{0.866+j0.5}{V}_P+y_t{V}_S$$
Representing these equations in standard nodal admittance matrix form yields the final correct option.