= Solution
For a <Cartesian second-rank tensor>, write
$$
P_{ij}=\frac12(P_{ij}+P_{ji})+\frac12(P_{ij}-P_{ji}).
$$
The first term is a <symmetric second-rank tensor> and the second is an <antisymmetric second-rank tensor>. Taking the <trace> of the symmetric term and subtracting it gives the <scalar, symmetric-traceless and axial tensor decomposition>:
$$
\boxed{P=\frac13P_{kk},\qquad S_{ij}=\frac12(P_{ij}+P_{ji})-\frac13P_{kk}\delta_{ij},\qquad A_k=\frac12\epsilon_{kij}P_{ij}.}
$$
Here $S_{ij}=S_{ji}$ and $S_{ii}=0$. To check reconstruction of the antisymmetric part, contract the <Levi-Civita symbols>:
$$
\epsilon_{ijk}A_k=\frac12\epsilon_{ijk}\epsilon_{klm}P_{lm}=\frac12(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})P_{lm}=\frac12(P_{ij}-P_{ji}).
$$
Thus $P_{ij}=P\delta_{ij}+S_{ij}+\epsilon_{ijk}A_k$, and the displayed contractions also prove uniqueness. The vector associated with the antisymmetric part is an <axial vector> under reflections, and an ordinary vector under proper rotations.
For the specified <isotropic tensor> law, contract the <Kronecker deltas> first:
$$
P_{ij}=\alpha\delta_{ij}T_{kk}+\beta T_{ij}+\gamma T_{ji}.
$$
Write $T_{ij}=T\delta_{ij}+W_{ij}+\epsilon_{ijk}V_k$, where $T=T_{kk}/3$ and $W$ is symmetric and traceless. Transposition preserves its scalar and symmetric-traceless parts and reverses its antisymmetric part. Therefore
$$
P_{ij}=(3\alpha+\beta+\gamma)T\delta_{ij}+(\beta+\gamma)W_{ij}+(\beta-\gamma)\epsilon_{ijk}V_k.
$$
Uniqueness of the decomposition gives the <isotropic elasticity on tensor components> relations. Provided their respective coefficients are nonzero, the requested inverses are
$$
\boxed{T=\frac{P}{3\alpha+\beta+\gamma},\qquad W_{ij}=\frac{S_{ij}}{\beta+\gamma},\qquad V_k=\frac{A_k}{\beta-\gamma}.}
$$
If one coefficient vanishes, that strain component has zero stress response: the corresponding stress component must be zero for a solution to exist, and that strain component is not uniquely recoverable. Isotropy by itself does not guarantee invertibility.
If $T_{ij}$ is symmetric then $V=0$, hence $A=0$ and the stress is symmetric. Equivalently, direct transposition gives $P_{ij}-P_{ji}=(\beta-\gamma)(T_{ij}-T_{ji})=0$. Restricting to this symmetric-strain case gives
$$
\boxed{P_{ij}=\lambda\delta_{ij}T_{kk}+\mu T_{ij},\qquad\lambda=\alpha,\quad\mu=\beta+\gamma.}
$$
This final two-parameter expression uses the symmetry assumption from the preceding request. For a general nonsymmetric array, the transpose term cannot be absorbed into the same coefficient: it acts with $\beta+\gamma$ on the symmetric-traceless part and with $\beta-\gamma$ on the antisymmetric part. The stated coefficient $\mu$ is the coefficient of $T_{ij}$ itself; if one instead writes the conventional elastic law with $2\mu$ multiplying the strain, that shear-modulus convention is half this coefficient.
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